NRLF 




311 



IN MEMORIAM 
FLOR1AN CAJOR1 





SIR ISAAC NEWTON (1642-1727) 



PRACTICAL ALGEBRA 



FIRST YEAR COURSE 



BY 



JOS. V. COLLINS 

PROFESSOR OF MATHEMATICS, STATE NORMAL SCHOOL 
STEVENS POINT, WISCONSIN 



NEW YORK .;. CINCINNATI : CHICAGO 

AMERICAN BOOK COMPANY 







COPYRIGHT, 1908, 1910, BY 
AMERICAN BOOK COMPANY. 

ENTERED AT STATIONERS' HALL, LONDON 



COLLINS'S FIRST YEAR ALGEBRA. 
W. P. 2 






CAJORf 



PREFACE 

THIS book has been prepared to meet the demand for a first 
year course by those high schools which make the study of the 
more advanced parts of algebra an elective. As its title indicates, 
the book aims to increase the practical value of elementary alge- 
bra, both in the ordinary and in the cultural or educational sense. 
It is written as well for those who do not go to colleges as for 
those who do. 

A wave of interest in mathematical teaching has swept over 
the United States during the last six or seven years, probably re- 
ceiving an impetus from the Perry movement in England. The 
many periodicals and pamphlets published by educators and edu- 
cational organizations interested in the reform movement have 
been carefully studied by the author, and suggestions from them 
have been freely used. 

The plan hitherto followed largely in America has been to keep 
arithmetic, algebra, and geometry separate. In foreign countries, 
on the other hand, the aim has been to make each of these 
1 tranches aid as much as possible in the study of the others. Per- 
haps the three most prominent features in the movement for 
reform in mathematical teaching have been, first, the attempt to 
make the work more practical ; second, the better correlation of 
the three branches, arithmetic, algebra, and geometry, with one 
another and with other branches, especially with science ; and 
third, the bringing down into the elementary course of the sim- 
plified elements of the higher mathematics. All of these features 
the author has sought to embody in his treatment, and a consider- 
able number of new practical problems have been introduced to 
apply these new ideas. 



4 PREFACE 

In developing a new topic the common plan is to follow the 
order, (1) general theory ; (2) several examples ; (3) exercises. 
In this book the following plan is pursued: (1) one example; 
(2) exercises, with suggestions, when needed ; (3) the making of 
a rule by the student based on his knowledge gained in solving 
the exercises ; (4) more exercises, following the rule ; (5) miscel- 
laneous exercises after sets of similar exercises. Evidently this 
plan follows the usual scientific order of induction, generalization, 
deduction. Or, stated pedagogically, each topic is presented fol- 
lowing the order: presentation, comparison, generalization, appli- 
cation. In the making of the rules some training is acquired in 
an important kind of composition in which conciseness, accuracy, 
and completeness are essential. Thus, the student gains strength 
both in mathematics and in English. 

A feature to which attention is directed is the proving and 
checking which runs all through the treatment, and which obvi- 
ates the necessity of an answer book for the student. Just as the 
bookkeeper has his double-entry, and the skilled calculator his 
checks, which he invariably uses, so the student should have his 
tests to apply to his answers. Mathematics is the one subject in 
which checking is feasible. Doing it systematically may pro- 
foundly affect the pupil's mental habits. 

The free introduction of what have been called informational 
problems, if it cannot be dignified with the name correlation, at 
least makes the algebra circle of knowledge touch other subjects 
in the curriculum and the world outside. These problems are 
even better adapted for algebra than for arithmetic, since simple 
relations of the numbers are introduced in the statements. Such 
problems are not hard to solve, and they give training in dealing 
with large numbers. 

In his treatment of graphs, the author believes that the simple 
presentation and specific directions will clear the subject of its 
usual difficulties for young learners. 

Other features of the book are: great clearness in presentation; 
the avoidance of difficult and complicated exercises; the abridg- 



PREFACE 5 

ment of the number of cases or kinds of problems that have to be 
learned; the constant use of the axioms in the solution of equa- 
tions; the presentation of factoring, and the deferring of the 
harder cases to the latter part of the course ; as well as the gen- 
eral conservatism in the arrangement of topics. 

Attention is directed to the Suggestions to Teachers on the next 
page. 

The author is indebted to Professor H. E. Cobb of Lewis Insti- 
tute, Chicago, for reading the chapter on practical applications 
and offering suggestions for its revision; to Miss Theresa Moran 
of the Stevens Point, Wis., High School for reading and criticiz- 
ing the whole manuscript ; and to Professor Robert E. Moritz, 
head of the department of mathematics of the University of 
Washington, Seattle, as well as to Mr. S. B. Todd of the Ameri- 
can Book Company, for valuable suggestions. In addition the 
following have read the proofs and have offered suggestions : Pro- 
fessor B. F. Finkel of Drury College, Springfield, Mo., editor of 
the American Mathematical Monthly; Professor W. H. Williams 
of the State Normal School, Platteville, Wis.; Miss Margaret 
French of the South Division High School, Milwaukee ; and Miss 
Mary C. Nye of the Blaine High School, Superior, Wis. 

STATE NORMAL SCHOOL, 
Stevens Point, Wis. 



SUGGESTIONS TO TEACHERS 

1. A Shorter Course Outlined. The number of problems in this 
algebra is exceptionally large. On this account some teachers 
may find it desirable not to require their classes to solve them 
all. To meet this condition, an abridgment of the regular drill 
work is provided by placing a star (*) before the first example 
of sets of problems that may be omitted without sacrifice of a 
sufficient amount of illustration of essential principles. Like- 
wise a star is used to mark some of the less important topics 
that may be omitted. By the omission of the starred topics anu 
sets of problems, a briefbut practical course will be provided for 
one year's work. 

2. Verifying, Checking, and Proving. It should be noted that 
the plan of proving results followed has the effect of greatly 
increasing the number of exercises. The verification of answers 
to all equations and equation problems is advised ; but checking, 
using figures for letters, can be easily overdone ; teachers should 
exercise judgment here, lest they retard the student's progress. 
Proving in division and factoring is urged whenever there is 
danger of error. This course keeps up a valuable review of 
multiplication and addition. 

3. Drill in Applying the Axioms. It is not necessary to em- 
phasize to experienced teachers the importance of persistent drill 
in applying the axioms in the solution of equations. Inexperi- 
enced teachers, however, very often follow the custom of nearly 
all authors of not referring again to this reasoning after it has 
been presented once, and in so doing they make a serious mistake. 
This book constantly asks for the reasons underlying operations 
in equations. 

4. Mental and Written Work. To be well balanced. Many 
problems throughout can be solved mentally, especially in reviews. 



TABLE OF CONTENTS 

(See Index at the end of the book for special topics) 

CHAPTER PAGE 

I. THE ALGEBRAIC NOTATION ....... 9 

Algebraic Numbers presented in the Arabic Notation . . 12 

Principles and Definitions ....... 24 

Addition and Subtraction ....... 30 

Symbols of Aggregation 40 

Multiplication 45 

Division 52 

Integral Equations 9, 37, 61 

Historical Note .64 

II. THEOREMS, FACTORING, LOWEST COMMON MULTIPLE . . 66 

Theorems in Multiplication and Division .... 66 

Factoring 74 

Quadratic Equations by Factoring .... .90 

Lowest Common Multiple 92 

III. FRACTIONS 96 

Highest Common Factor . .98 

IV. SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY . . 122 

V. SIMULTANEOUS EQUATIONS. ' Two UNKNOWN QUANTITIES. . 154 

Simultaneous Equations. Three Unknowns . . 168 

VI. GRAPHS AND GRAPHICAL SOLUTION OF SIMPLE EQUATIONS . 172 

VII. MISCELLANEOUS TOPICS .... 185 

Simple Quadratic Equations ... 186 
Literal Equations ...... 

Generalization ....... . 193 

Indeterminate Equations ..... . 193 

Negative Solutions 193 

Supplementary Factoring 194 

Historical and Biographical Notes .... . 199 

VIIL INVOLUTION AND EVOLUTION 202 

7 



8 TABLE OF CONTENTS 

OHAPTEB PAGE 

IX. APPLICATIONS OF ALGEBRA 220 

Applications in Arithmetic 220 

Applications in Geometry 230 

Applications with Squared Paper 236 

Applications in Physics 240 

Formulas 242 

Manual Training and Domestic Science .... 246 

X. FRACTIONAL EXPONENT QUANTITIES AND RADICALS . . 250 

Radical Expressions 257 

Imaginaries . 270 

Equations containing Radicals 270 

Historical Notes 273 

XI. QUADRATIC EQUATIONS 276 

Simultaneous Quadratics 293 

Algebra used in Trigonometry 298 

INDEX ... ........ 299 



PRACTICAL ELEMENTARY ALGEBRA 

CHAPTER I 

THE ALGEBRAIC NOTATION 

1. Algebra is a study of numbers. It differs from arithmetic in 
one way by using letters as well as figures to denote its numbers. 

The student should distinguish clearly between a number itself 
and the mark that stands for it. One definition often given is 
that a number is a ratio, or a quotient of one magnitude divided by 
another of the same kind. For example, the ratio of the length 
of a room to a unit length (foot, or meter) is a number. To denote 
this number we can use a word, or figures, or a letter. The fol- 
lowing examples will show how letters can stand for numbers. 

1. There are three times as many girls as boys in a school con- 
taining 88 pupils. How many are boys ? 

ARITHMETICAL SOLUTION 

A certain number = the number of boys in the school ; 
then 3 times that number = the number of girls in the school ; 

and 4 times that number the number of boys and girls together. 

Hence, 4 times that number = 88, 
and that number = \ of 88, or 22. 

Thus, the number of boys in the school is 22. 

In the algebraic solution of this problem, which follows, instead of writing 
" a certain number," or "that number " each time, we abbreviate by writing 

x for this phrase. 

ALGEBRAIC SOLUTION 

Let x = the number of boys in the school ; 

then 3 x = the number of girls (3 x means 3 times #), 

and 4 x = the number of boys and girls together. 

88 = the number of boys and girls together. 
Hence, 4 x = 88 ; 

then x = of 88 = 22, the number of boys. 

9 



JO THE -ALGEBRAIC NOTATION 

PROOF. If there are 22 boys in the school, and three times as many girls 
as boys, then there are 66 girls. Altogether there are the sum of 22 and 66, 
or 88 pupils, which is the number the problem says there are. Hence, the 
answer is right. 

2. There are 7 sheep for every lamb in a pasture, and 96 sheep 
and lauibs altogether. How many lambs are there ? Work by 
both arithmetic and algebra, following the models on p. 9, and 
prove answer. 

3. There are three packages, the second of which weighs twice 
as much as the first, and the third three times as much as the first. 
When put together on the scales they balance a 24 Ib. weight. 
What is the weight of each package ? 

SOLUTION. Let x = the number of pounds in weight of first package ; 
then 2 x the number of pounds in weight of second package, 

and 3 x = the number of pounds in weight of third package. 

Then x + 2 x + 3 x = 24, the sum of the weights of all, 
or, 6 x 24, by adding x, 2 #, and 3 x ; 

then x = of 24 = 4, number of pounds in 1st package, 

2 x = 2 x 4 = 8, number of pounds in 2d package, 
3 x = 3 x 4 = 12, number of pounds in 3d package. 

PROOF. 4 + 8 + 12 = 24 ; also 8=2x4, and 12 = 3 x 4. 

4. Three persons together weigh 350 Ib. The second weighs 
twice as much as the first and the third 4 times as much as the 
first. What is the weight of each ? Prove answer. 

5. Two men, each working for $ 1 a day, received $ 132 for their 
labor. A worked 11 days every time B worked 1 day. How many 
days did B work and how many did A work ? Prove answer. 

SUGGESTIONS. Let x = the number of days B worked. Then how many 
dollars did B receive? Ans. $x. If B worked x days and A worked 11 
tinies as many clays, how many days did A work ? Ans. 11 x. Then how 
many dollars did A receive ? Now how many dollars does the problem say 
x + 1 1 x equals ? 

6. A horse and saddle cost $ 150. If the horse costs 5 times 
as much as the saddle, what is the cost of each ? Prove answer. 



THE ALGEBRAIC NOTATION 11 

7. Of 240 stamps collected by a boy and his sister, the boy col- 
lected 4 times as many as his sister. How many did each collect ? 
Prove. 

8. Twice a certain number added to 5 times the number equals 
98. What is the number ? Prove answer. 

9. A farmer raised a certain number of bushels of wheat, 4 
times as much corn, and 3 times as much oats. If there were in 
all 4000 bu. of grain, how many bushels of each kind did he raise? 

10. A manufacturer who has 720 men working for him wishes 
to put a certain number in one building, 3 times as many in a 
second, and twice as many in a third building as he puts in the 
second. How many men must he put in each ? 

11. The area of the land surface of the earth in round num- 
bers is 51,200,000 square miles. The area of the rest of the 
earth's land surface is 7 times that of North America. What is 
the area of North America in square miles ? Prove answer. 

12. The total population of the Philippines in round numbers 
is 7,634,000. Of this number there are 10 times as many civilized 
people as there are uncivilized people. How many are there of 
each ? Prove answer. 

13. In 1900 when the population of the United States in even 
thousands was 76,303,000, there were 3 times as many people 
living in small cities (under 30,000) and in the country as in large 
cities. How many lived in large cities ? Prove answer. 

14. The Panama Canal will be 46 miles long. Of this distance 
the lower land parts on the Atlantic and Pacific sides will together 
be 9 times the length of the Culebra Cut, or hill part. How many 
miles long will the Culebra Cut be ? Prove answer. 

15. The value of all other property in the United States in 1900 
was about 5.8 times the value of manufactured articles, and the 
total wealth was estimated to be 88.5 billions of dollars. What 
was the number of billions of dollars in the value of the manufac- 
tured articles ? Prove answer. 



12 THE ALGEBRAIC NOTATION 

r 

2. The value of a letter, as we see from the preceding article, 
does not change throughout the solution of any problem. But 
the same letter, as x, may have different values in different prob- 
lems. This is not the case with figures. Thus, 4 always stands 
for four. 

3. Symbols of Omission or Continuation. Dots or periods are 
commonly used in algebra to indicate that figures or letters not 
written are to be understood. They are read " and so on to" or 
(t and so on." 

Thus, 1, 2, 3, 4, ... 20 is read "one, two, three, four, and so on 
to twenty " ; while 1, 2, 3, 4 ; is read " one, two, three, four, and 
so on indefinitely." 

4. Algebra and arithmetic are unlike, as explained in 1, in 
that algebra uses letters as well as figures to denote numbers. 
They are unlike also in another important respect, namely, arith- 
metic has but one series of numbers, while algebra has two series, 
extending in opposite directions from zero. 

Arithmetic series, 0, 1, 2, 3, 4, 5, 6, 

Algebra series, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 

The height of a tree, the age of a man, the number of silver 
dollars a man has in his pocket, etc., are all arithmetical numbers 
because their lowest value is 0. The number of dollars a man 
has to his credit on the books of a bank, a temperature, a date 
A.D. or B.C., etc., are all algebraic numbers, because such num- 
bers can lie on either side of zero. There are vastly more con- 
crete numbers that are arithmetical than are algebraical. Examine 
the room you are in for examples of the two kinds of numbers. 

5. Algebra may be described as that branch of mathematics which 
uses letters as well as figures in the study of numbers, and separates 
its numbers into two opposite kinds. 

6. Addition of Algebraic Numbers. Addition is literally " put- 
ting together." One gets a good idea of algebraic addition by 
thinking of a business man receiving during a day pieces of paper 



ADDITION 13 

representing money, as checks, bank notes, promissory notes, due 
bills, and other evidences of credit and debt. These are "put 
together " on a pile. The sum is found by adding the debts and 
credits separately, subtracting the less amount from the greater, 
and marking the result credit or debt according as the credit or 
debt sum is the greater. To distinguish credits from debts we 
will mark the credits with a small + sign and the debts with a 
small " sign. Thus, + 9^ means a credit of 9^, and is read "plus 
9 cents " ; similarly $ ~3 is read " minus 3 dollars." 



This problem evidently calls for the sum of a credit of 
another credit of 5^, a debt of 12 X, another debt of 15^, and a 
credit of 29 

SOLUTION. +6^ + +5^ + +29^ = +40^ ; -12^ -f -15 / = -27^ ; 

+40^ + -27 ?=+13f'. Ans. 

EXPLANATION. In the solution, the addition of all the credits gives a total 
credit of 40^, and the addition of all the debts gives a total debt of 
Then the sum of a credit of 40^ and a debt of 27^ is a credit of 



CHECK. The correctness of the answer just found can be "checked," or 
tested, by adding the items in the order in which they come. 

Thus, +6^ + +5/ = +11^ ; +11/' -f -12^ = ~l? ; ~1/ + ~15^ = - 
4- +29 / = +13^, or the same answer as before. 



Solve the following problems in the same way the problem just 
given was solved, and check the answer in the same way that 
answer was checked : 



2. + 12 + -3 + -T^ + + 15^ + -2=? Ans. 

3. 

4. 

5. 

6. 



14 THE ALGEBRAIC NOTATION 

In the following problems set down the several items with -f- 
signs between them, marking the credits with small + signs and 
the debts with small " signs, and find the sum : 

7. George has 16 / in his money box and 10^ in his pocket. 
He owes his brother John 7 '$ and a confectioner 14^. William 
owes him 10^. How much money has he? 

8. William, who has 25^ in money in his pocketbook, owes a 
grocer 8^, a pieman 4^, his sister 3^, and a playmate 15^. His 
brother James owes him 17^. How much money will he have left 
if he collects what is due him and pays his debts? Solve as in 
the preceding. 

9. A man who has $300 in the bank and $125 in his pocket 
owes A $550 and B $1000. He has due him from one party 
$400 and from another $640; but he owes as surety on a note 
$190. Find and mark how he stands financially as in the pre- 
ceding problems. 

10. A ship starting from 1 north latitude goes one day 2 
north, the next 3 north, a third 4 south, a fourth 3 south. If 
north latitude and north sailing are marked + and south latitude 
and south sailing ~, what latitude is the ship now in and how is 
the latitude marked ? Solve as in the preceding problems. 

11. A thermometer, which stood at 4 P.M. at 70, and which by 
8 o'clock had fallen 14, and by midnight 8 more, and by 4 A.M. 
3 more, rose from 4 o'clock to 8 o'clock 12, and from 8 o'clock 
to 12 o'clock noon 25. If above zero and upward movement are 
marked + , and downward movement and below zero ~, what is 
the position of the mercury at noon of the second day and how is 
it marked ? Solve as in the preceding problems. 

12. A man deposits in a bank $ 600. He then draws checks at 
different times for $60, $19, $25.50, $10.19. Later he deposits 
$75 at one time and $32.12 at another. Afterwards he draws 
checks for $45.14 and $13.72. How does his account then stand 
on the bank's books ? Mark deposits + and checks ~, and solve 
as in the preceding. 



SUBTRACTION 



15 



7. Subtraction of Algebraic Numbers. Subtraction in algebra is 
the process of finding a number which added to one of two 
given numbers (the subtrahend) produces the other (the minuend). 
As in arithmetic, the result is called the difference. 

1. $+12 --$+7 = ? This problem, it is plain, risks for the 
difference between a credit of $ 12 and a credit of $ 7. 

SOLUTION. 8 + 12 - $+7 == 8+5. Ans. 

EXPLANATION. An algebraic number must be found which being added to 
the subtrahend 8+7 gives the minuend $ + 12. Clearly $+5 added to $+7 
equals $+12. Hence $+5 is the difference sought. 



4-4- i A it 4- O "" d O K < + ^5 O Q *? 

. ^40^ _>?' = : 5. $> o <j> o = r 

Ex. 5 evidently asks for the difference between a credit of 8 <> and a debt 
of 83. 

SOLUTION. $ +6 - $-3 = 8 +9. .4ns. 

EXPLANATION. A number must be found which being added to 8 ~3 gives 
the minuend 8+6. Now it takes 8 + 9 added to 8 ~3 to make 8+6. Hence the 
difference sought is 8 +9. Thus, the difference in fortunes of two boys, the 
first of whom is worth 8 6 and the second of whom 
is 83 in debt, is 8 9. The first boy is 89 better off 
than the second boy. 

Solve the following four problems in the 
same way as Ex. 5, and explain the results : 

6. $+10-$-4 = ? 7. +25? --lot = ? 
8. 22t--16t = ? 9. 12- -5==? 

Thus, a thermometer stood 12 above 
zero, and the day before 5 below zero. 
What was the difference in temperature ? 

a. REMARK. It is to be understood when a problem asks " what is the dif- 
ference between two given numbers" that the number first named is the 
minuend and the other the subtrahend. 

10. What is the number of years' difference in time between 
26 A.D. (which we will call +26) and 10 B.C, ? 




16 THE ALGEBRAIC NOTATION 

11. The difference between two algebraic numbers is the num- 
ber of units of distance between them on the algebraic scale, marked 
+ or ~ to show their relative position. 

^6~^5~-4 -3 -2 -1 O +1 + 2 +3 +4 +5 +6 

We represent the algebraic scale here in both horizon- 

_ -) O 

tal and vertical positions. The numbers are said to 
be above zero and the numbers below zero, as on a 
thermometer scale. 

No matter what position the scale has, from towards 
the + numbers is called the + direction, and from to- 
wards the - numbers is the ~ direction. If one number 
lies in the + direction from another, it is said to be [_- 4 
higher up in the scale than the other number. Thus, +2 h - 
is higher in the scale than ~4. A man who has $2 is 
better off than another who is in debt $4. 

If now the minuend is higher up in the scale than the subtra- 
hend (as it always is in arithmetic), the difference is marked + ; 
if the minuend is lower down in the scale than the subtrahend, 
the difference is marked". 

Thus, a difference in algebra tells two things : first, the number 
tells how far apart the minuend and subtrahend are ; and, second, 
the sign + or ~ tells whether the minuend is above the sub- 
trahend or below it. 

Find the difference in the following problems by using the dia- 
gram. See that the answer is marked with the proper sign. The 
student will find it helpful to think of the numbers as denoting 
degrees on the thermometer. 

12. What is the difference between 12 above zero and 4 below 
zero, that is, between +12 and ~4 ? Ans- +16. Because + 12 and 
~4 are 16 units apart and the minuend is above the subtrahend. 

13. +16-+10 = ? 14. +2--2 = ? 15. +11- +15 = ? 
16. ~8 -~10 = ? 17. -7- +6 = ? 18. ~6- ~4=? 
19. Find the difference in temperature between January 5, when 
it was 6 below zero, and January 4, when it was 13 below zero. 



SUBTRACTION 



17 



20. What is the difference in time between 1600 A.D. and 500 
B.C., if A.D. is marked + and B.C. is marked ~? 

21. Go over the problems from 12 to 20, working them this time 
by the method of finding a number which added to the subtra- 
hend will give the minuend, as explained at the beginning of this 
article. This time the student should think of the numbers as 
denoting dollars or cents. The answers are always the same, 
whichever method is used. 

22. The method with the scale which was used in Ex. 12-20 is 
more satisfactory as an explanation of subtraction, but the method 
introduced first is better for practical use. We must know how 
to add in any event, and we can always make subtraction depend 
on addition. Hereafter, therefore, we will use the rule stated in 
the next paragraph. 

23. To subtract, find a number which added algebraically to the 
subtrahend gives the minuend. 

Or, conceive the sign of the subtrahend to be changed, and add 
the result algebraically to the minuend. 

Subtract in following : (check, thinking of debts and credits) 



+ 



12 



REASON. Because -5 and +12 added give +7. 
CHECK. A credit of $ 12 added to a debt of $ 5 gives a credit 
of 8 7. 



25. 


+7 


26. 


-11 


27. 


-15 


28. 







+2 




"22 




+7 




2^ 


29. 


+25 


30. 


-48 


31. 


+40 


32. 


-49 




+31 




-39 




-51 




+4 

^^^^ ^^ 


33. 


-6 


34. 


-27 


35. 


+25 


36. 


+35 




-16 




-6 ' 




-17 




+46 



37. +25-+19. 38. -30 --36. 39. -12- -40. 40. ~6 - +36. 
41. -24-0. 42. ^50 --68. 4.3. + 169-~75, 44. -324-+125, 



18 THE ALGEBRAIC NOTATION 

45. What is the difference between the fortunes of two men, 
the first of whom owes $3000, and the second owes $3335 ? 

46. What is the difference in latitude between Santiago de Cuba 
20 north and Buenos Ayres 34| south, calling north latitude + ? 

47. What is the difference in longitude between Berlin 13 24 'E. 
and New York 74 W., if east longitude is taken as + ? 

48. What is the difference between the average July tempera- 
ture + 65 and the average January temperature ~8 at. St. Vincent, 
Minn. ? 

In the following problems + denotes A.D. and ~ B.C., as in the 
Standard Dictionary. 

49. The reign of Augustus Caesar over the Roman Empire 
ended with his death in +14, having begun in ~31. How long was 
he sole ruler ? 

50. The Roman historian Livy died in the year +17, having 
been born in the year ~59. How long did he live ? 

51. The Roman poet Horace died in the year ~8, having been 
born in the year ~65. What was his age when he died ? 

52. Columbus discovered America in +1492. The earliest 
accounts we have of the sailors of the ancient world go back to 
about ~1500. How long a period elapsed from the earlier to the 
later date ? 

53. Greece became a Roman colony "146, two hundred fifty- 
three years after the philosopher Socrates drank the cup of 
hemlock. In what year did Socrates die ? 

SUGGESTION. Mark 253 as -'-253 and subtract it from ~146. 

*54. Pope Gregory XIII reformed the calendar in +1582. The 
Council of Nice set Easter +325. Find the difference in time. 

NOTE. Probs. 49-54 are indefinite. In Ex 49 from Jan. 1, -31 to 
Dec. 31, +14 is +45 yr. But from Jan. 1, -31 to Jan. 1, +14 is 1 yr. less. 

Find interval from Sept. 10, -20 to June 15, +50. Solution : Sept. 10, 
-20 is 19 yr. 3 mo. 20 da. before era, and June 15, +50 is 50 yr. 6 mo. 15 da. 
after it. Ans. 68 yr. 9 mo. 5 da. Answer may vary 1 day either way. 



ADDITION AND SUBTRACTION 19 

55. The Swiss defeated the Austrian Hapsburgs at the battle 
of Morgarten in +1315. The Greeks defeated the Persians at the 
battle of Thermopylae "480. What number of years intervened 
between these battles ? 

56. Titus destroyed Jerusalem in the year + 70. Probably the 
epoch of greatest glory in Jewish history was at the time Solomon 
dedicated the temple at Jerusalem ~970. How many years elapsed 
from the earlier to the later date ? 

57. The Roman Empire, as commonly stated, fell in the year 
+ 476. This was 622 years after the destruction of Carthage. In 
what year was Carthage destroyed ? 

REMARK. Ex. 54, p. 18 (marked with a star), and Ex. 55-57 following, 
are to be omitted in the Shorter Course. See 1, p. 6. 

8. New Marks for the Series. The sign + is always understood 
when no series sign is written. Thus 4 means + 4. Now 

+ 7 1 = + 7 ^. (Since adding 7 f to gives 7 $. ) 
3^ = ~3^. (Because the sum of a debt of 3^ and a credit 

of 3^ = 0, 7.) 

Thus, + 7 can replace + 7, and 3 can replace ~3. But 
can be dropped, as it has no value. Dropping 0, we have, 

= + 7and -3 = ~3. 



Thus, the small + and " signs written above the line of writ- 
ing, and used to mark the series to which numbers belong can be 
replaced by large + and signs. It is understood when no sign 
is written that + is understood. 

When algebraists desire to indicate that a number belongs to 
the positive (+) or negative ( ) series, they put it with its sign 
in parenthesis. Thus, 6 + (4) denotes that 4 in the negative 
series is to be added to 6. 

9. Addition and Subtraction of Algebraic Numbers. By means of 
the series sign just described, an example in subtraction can be 
changed into one in addition. 



20 THE ALGEBRAIC NOTATION 

Thus, 6 4, that is, 6 (+4), equals 6 +(4), since each 
equals +2. 

Algebraists commonly regard such a problem as 

6 + 9-3 + 4-12 
as one in addition, thinking of it as written 



SOLUTION. The sum of the + numbers is -f 19 : the sum of the num- 
bers is - 15 ; + 19-f (- 15) = + 4. 

CHECK. (See Explan. G.) 
6 + 9 = 15; 15 +(-3) = + 12; +12+( + 4)= +16; + 16 + (- 12) = + 4. 

Add and check in the problems of this article as in the model 
just solved. 

1. 12-8 + 6-7 + 15. 2. 6-7 + 15 + 2 + 5-9. 

3. _3 + 4_9_i2 + 15. 4. -15 + 12-10-8+3-12. 

5 . 19 _ c + 21 + 65 -71. 6. 78-44 + 39-25 + 17. 

7. On a business man's table there lay at the end of a day's 
business the following pieces of paper: three ten-dollar bills; a 
note in his favor for $25; one due bill against him of $12, and 
a second of $ 23 ; two checks in his favor, one of $ 19 and another 
of $22; and a memorandum of a note given to the bank in its 
favor for $40. What is the sum of these items ? 



8. The price of wheat in January was 75 ^. During February 
it rose 7j^; then in March it fell 12^; in April it rose 8/, and 
in May 10^; and in June it fell 2^. What was the price at 
the end of June ? Mark upward movements + and downward 
inside parentheses. 

9. Three toy balloons are fastened together, the first having 
an upward pull of 2 oz., the second of 3 oz., and the third of 1 oz. 
To them are fastened weights of 21 oz. and 1^ oz., respectively. 
If upward pull is marked + and downward , what is the sum 
of these forces ? 



MULTIPLICATION 21 

10. If an elevator starts on the ground floor and goes up 30 ft., 
then down 15 ft., then up 45 ft., then up 18 ft., and last of all 
down 33 feet., where is it then, counting from the ground floor ? 

11. A man rows 5 mi. an hour in still water and the current of 
the river he rows in is 3 mi. an hour. If motion downstream is 
marked -f- and upstream -, what is the man's rate of progress 
when he rows upstream ? What is it when he rows downstream ? 

10. Multiplication in algebra is like multiplication in arith- 
metic, aside from the question of the sign of the product. This 
matter we will now consider. (In this article for x read " times.") 



SOLUTION. Once 7 ? = 7 ^ ; twice 7 ^ = 14 ^, etc.; 6 x 7 / = 42^. Or, 
6 credits of 7 ^ each = a credit of 42 ^. 

A positive number times a positive number gives a positive product. 



2. 6 x-7^=? 

SOLUTION. Once 7 $ = 7 ^ ; twice 7 ^ = 14 ^, etc. ; 6 x 7 



Or, 6 debts of 7 ^ each equal a debt of 42 p. 
A positive number times a negative number gives a negative product. 

3. -2x7^=? 

SOLUTION. 2x7^ = 14 ^ 

_ Notice that the multipliers decrease by 1 

ft 7 <? " ft <? as y u down, anc " 1 tne products decrease 
1 _~ by 7^ each time. The last multiplication, 

-2x7^=^14^ -2x7^ =-14^, shows that, 

A negative number times a positive number gives a negative product. 



4. _2x- 

SOLUTION. 2 x--7^= 14^ (By 2 above.) Notice as in 3 above, 

lx--7^= 7^ the multipliers decrease by 1 each time, 

Ox 7 ? = Op but that the products increase by 7 ^ 

lx 7^=+ 7^ each time. The last multiplication, 

2x 7^=+14^ _2x 7^=+14^, shows that, 

A negative number times a negative number gives a positive product. 



22 THE ALGEBRAIC NOTATION 

5. Combining the results in italics in the preceding paragraphs, 
we have this general rule : 

In multiplication two like signs give -f in the product, and two un- 
like signs give . 

a. The result under 4, where a negative number times a negative number 
gives a positive product, is hard for beginners in algebra to understand. It is, 
however, not unlike the grammatical rule which says that "Two negatives 
make an affirmative." Thus not unlike means like. 

6. The rules for signs can be pictured to the eye in the fol- 
lowing way. Using an algebraical series of numbers on a hori- 
zontal line, we observe that in lxH-3 = 3 (by the rule 
in 3 p. 21), the multiplier 1 turns the multiplicand + 3 around 
through an angle of 180 to the position - - 3. 

Eeasoning in the same 
way, 2 x 2 = + 4, 
because - - 2 as multi- 
plier first doubles multi- 
plicand --2, giving- 4, 
and then reverses the 
direction of 4. 
A positive multiplier takes the multiplicand so many times in its 

own direction, and a negative multiplier takes the multiplicand so 

many times in reversed direction. 

In solving the following problems, use the rule given in 5 above, 
to get the sign of the product : 

7. 3x-12 = ? 8. -7x-4 = ? 9. +5x+9 = ? 

10. -15x6 = ? 11. -15x-12=? 12. +8x-13 = ? 

13. -40x-20 = ? 14. _7x+16 = ? 15. -8x+13 = ? 



....-7-6-5-4-3-2-1 + 1 + 2-1-3-1-4 +54-6 + 7 




16. Make a rule now in your own words for the sign of the 
product when there are two factors multiplied together. 

17. 3x 4x 5 = ? SOLUTION. 3 x -4 =- 12 ; - 12 x - 5 = -f 60. 



DIVISION 23 

18. -6x-2x-7 = ? 19. 4x5x-12 = ? 

20. 6x7x2 = ? 21. -2x-2x-3x-4=? 

22. 5x3x--5x6x2x 1 = ? 

23. -lx--lX--lX--lX--l-=? 

24. 2x-2x-3x-4xl X+5-? 

25. Answer now the following questions : When there are no 
minus factors, what is the sign of the result ? When there is one 
minus factor, what is the sign of the product ? When there are 
two minus factors ? When there are three minus factors ? When 
there are four minus factors ? When there is an odd number of 
minus factors ? Will the following rule cover every case ? When 
there is an odd number of minus factors the product is negative ; 
otherwise the product is positive. 

26. 3x4x-2=? 27. 6x-6x-l=? 

28. -2x3x-lx-5=? 29. -2x-3x-7 x-l = ? 

30. 3x5x-2x-6x-4 = ? 31. -lx-3x-4x -5x-2=? 

11. Division of Algebraic Numbers. 

1. If +6x +3- + 18, then +18-=-+G=+3, or a + no. -f- + no. gives + quotient. 

2. If +6x 3 = 18, then 18 -=-+6= 3, or a no. -=- +no. gives quotient. 

3. If 6x +3 = 18, then 18-^ 6= +3, or a no. -=- a no. = + quotient. 

4. If -6x 3= +18, then +18-f- 6= 3, or a + no. -* a no. = quotient. 

5. From this we can see that the rule for division is much like 
that for the multiplication of two factors, viz., Like signs in divi- 
dend and divisor give +- in the quotient and unlike signs give . 



6. 24-=--6 = ? 7. _15-s--5 = ? 8. -16- + 4 = ? 
9. 64n--4 = ? 10. -32--8 = ? 11. -2.5-s-.o = ? 



12. -|-5--| = ? 13. !-= 1 = ? 14. 6.3-5-2.1 = ? 



24 THE ALGEBRAIC NOTATION 

12. Miscellaneous Exercise in the Four Operations. 
1. 6-(-8). 2. 4+6-11+3-17-5+19. 3. -8-(-14). 
4. 4x-7x-2x5. 5. -24-3. 6. -30- -6. 7. 4-12. 

8. If north latitude is marked + and south latitude -, what 
is the difference in latitude between two ships the first at 43 north 
and the other at 17 south latitude? Answer shows which direc- 
tion first is from second. 



9. A boy has 25$ in his bank, 15^ in his pocket, and another 
boy owes him 40^. He owes 60^. How much money has he ? 

10. On one day the mercury stood at 14 above zero and the 
day before it marked 13 below zero. What was the difference in 
temperature between the two days? 

11. What is the difference in longitude between Philadelphia, 
which is about 75 9' west, and Sacramento, which is 121 20' west, 
west being marked ? 

PRINCIPLES AND DEFINITIONS 

13. The Fundamental Laws of Operation in Algebra. These laws 
are assumed to hold true throughout algebra. It is easy to illus- 
trate their truth in special cases. 

1. Commutative or Order Law in Addition. The order in which 
numbers are added in algebra does not affect the sum. 

Thus, it is well understood that the order in which debits and 
credits are arranged makes no difference in the total. 

For example, 
$3+(-$5)+$7=(-$5 



.2. Associative or Grouping Law in Addition. The way in which 
numbers are grouped, or associated together, in adding them makes 
no difference with their sum. 



PRINCIPLES AND DEFINITIONS 25 

In algebra, numbers are associated together by means of 
brackets or parentheses. In a business house the different items 
of debt and credit can be grouped together in any way by means 
of rubber bands (which correspond to the parentheses), but the 
way they are associated makes no difference with the total value. 

Thus, ($3+ $-4) + $5= $3 + [$-4+ $5]. 

3. Commutative or Order Law in Multiplication. The order in 
which factors are arranged in a product can be changed without 
changing the product. 

Thus, + 4x--5=-5x+4; also 2x3x-4=--4x2x3. 

4. Associative or Grouping Law in Multiplication. TJie factors of 
a product can be arranged in anyway in groups without changing 
the product. 

Thus, (3x4) x -5 = 3 x (4 x -5). 

5. Distributive Law in Multiplication. In all forms of multipli- 
cation every part of the multiplicand must be multiplied by every 
part of the multiplier and the partial products added. 

Thus, 3 + 4 = 7 59 

2 + 6 = x 8 x!7 

6 + 8 56 413 

+ 18 + 24 59 

6 + 8+18 + 24 = 56 1003 

Had any partial product been omitted, the product would have 
been wrong. 

If the pupil will point to each figure named as he multiplies, 
he will be reminded of a postal clerk in a post office or mail car 
distributing mail. 

14. The Algebraic Notation. In the algebraic notation the 
absence of a sign between letters or between a figure and a letter 
indicates multiplication. 

Thus, 7x means 7 times a;; abc denotes the product of the 
numbers represented by a, b } and c. 



26 THE ALGEBRAIC NOTATION 

In arithmetic the absence of a sign between figures denotes 
addition. 

Thus, 34 = 30 + 4 ; 5 = 5 + . 

In algebra, as well as in arithmetic, numbers expressed in 
figures are understood to be written in the Arabic notation. 

Thus, in algebra 29 = twenty-nine. 

In algebra, as in arithmetic, x denotes multiplication. The 
period raised above its usual position is also sometimes used to 
denote multiplication. Thus, 6 7 = 6 X 7 = 42. 

An oblique line is frequently used to denote division. Thus. 

,, ; a 

a/o means a-t-o, or - 

b 

15. A quantity in algebra is a number. This number may be 
expressed by a single letter, or by a figure, or by any combination 
of figures, letters, and signs. 

Thus, 7, o, and 2a + 36 5c are each quantities. 

16. An exponent is a number written to the right of and higher 
than a quantity. When it is a positive whole number, it shows 
how many times the quantity is to be taken as~a factor. Ex- 
ponents are essentially a kind of shorthand. The real thing meant 
is not written and must be understood. 

Thus, 4 2 = 4 x 4 ; a 3 = a X a X cr; a?=x X x x x X x X x. When no 
exponent is written after a quantity, 1 is understood. Thus, a=a 1 . 

17. A power of a number is the product obtained by using 1 and 
the number one or more times as factors. The product of 1 by a 
number multiplied by itself is called its square, and of 1 and the 
number used three times as a factor, its third power, or cube. 

Thus, a 2 is a power of a, and is read " a square " or " a squared " ; 
a 4 is read " a fourth power," or simply " a fourth." 

18. A root of a number is one of its equal factors. One of 
two equal factors of a number is called its square root; one of 
three, its cube root; one of four, its fourth root; etc. 



PRINCIPLES AND DEFINITIONS 



A symbol to denote roots is the radical sign y'. Thus, V9 
denotes the square root of 9, or 3 ; Va denotes the cube root of 
a ; -\/V7, the fourth root of b. The figure above the radical sign 
is called the index of the root. 

19. A term is an algebraical expression, or quantity, the 
parts of which are not separated by + and - - signs ; as o ab. 

20. The degree of a term is the sum of the exponents of its 
literal (letter) factors. Thus 4 crfr 3 is of the iifth degree. 

21. Quantities are classified according to the number of terms 
they contain into monomials, consisting of one term, as 3ab~: bi- 
nomials, consisting of two terms, as 5m?i+4p 2 ; trinomials, con- 
sisting of three terms ; and polynomials, consisting of several 
(literally, many) terms. For convenience, however, quantities 
are often classified into two kinds : monomials, consisting of one 
term, &EL& polynomials, consisting of two or more terms. 

22. The recognized parts of a term are : (1) The numerical co- 
efficient, or simply the coefficient. (2) The literal part. 

Thus, in 6 a?b, 6 is the coefficient and a?b is the literal part or 
letter part. The coefficient shows how many times the literal part 
is taken. 

Thus, 6 a 2 b = a~b + a-b + a 2 b + a 2 b + a 2 b + a 2 b. 

In a broader sense a coefficient may be defined as either one of two factors 
into which a number can be resolved. Thus, in 9 cy, 9 is the coefficient of cy, 
cy of 9, 9 c of ?/, 9 y of c, etc. 

23. Similar terms are those which have the same literal part, 
that is, the same letters with the same exponents. Two similar 
terms can differ in signs and coefficients. 

Thus, -f- 6 x^y 1 and 11 x*y 2 are similar terras. 



28 THE ALGEBRAIC NOTATION 

24. The numerical value of an algebraical quantity is found by 
assigning definite values to the letters and simplifying the result. 

EXAMPLE. What is the numerical value of 3 a + 2 b 4, when 
a = 5,b=3? If a = 5, and 6 = 3, 3a+2 6-4=3x5 + 2x3-4 = 17; 

thus, 17 is the required numerical value of the given quantity. 

25. The definitions of many words familiar to the student in 
arithmetic are not given here. If the student does not know the 
meaning of any word used, let him look it up in the dictionary. 

26. The symbols of aggregation are the parenthesis ( ), brace -J }, 
bracket [ ], and vinculum " . They are used to indicate that 
all the terms within them are to be treated as one quantity, or 
that the operations indicated within come before those without. 

Thus, in 3 x (4 + 5), 4 and 5 must first be added, then the sum 
must be multiplied by 3. 

27. Order of Precedence of Operations. Without reference to the 
order in which the terms of a quantity are written, the following 
is the order in which the operations indicated must be performed, 
unless symbols of aggregation indicate that certain parts are to be 
considered as one quantity : 

1. In any term, power and root operations are to be performed 
first and multiplications and divisions afterwards. 

2. TJie operations of addition and subtraction come last. 

$} 
Thus, 2 x 3*=: 18 j 5x2 2 -2 3 -4 = 18; 3 + 4x5 = 23. 

Similarly, if a = 3, 6 = 2, c = 16, j 



becomes 15 x 3 x 2* - 3 x 2 x Vl6, (By 14) 

that is, 15 x 3 x 4 - 3 x 8 x 4, or 84. ( 16, 18) 

REMARK. In c x - the division, and in p -~ mn the multiplication is per- 

formed first. 3 x 4 H- 2 x 6 calls for the use of parentheses, 

since 3 x 4 + (2 x 6) =1, while (3 x 4 *- 2) x 6 = 36. 



PRINCIPLES AND DEFINITIONS 29 

3. Calculate 3 ab 2 c - 2 aW Vd + 2 c, when a = 1, 6 = 2, c = 2, 
and d = 4. 

SOLUTION. 3 & 2 c = 3 x 1 X 4 x 2 = 24 ; 2 a& 3 Vd = 2 x 1 X 8 x 2 = :]:> ; 
2 c = 4 ; then 24 -- 32 + 4 = - 4. 



4. Calculate 6 2 # 9 y 2 when a = 4, a? = 2, y = 3. Ans. 111. 

5. Calculate 3 a 2 4- 2 c# 6 3 when a = 5, b = 7, c = 15, a; = 3. 

6. Calculate 9 a~bc + 3 Vc - & 2 V& when a = 2, 6 = 4, c = 25. 

7. As stated before, the rules in 1 and 2 do not hold when 
symbols of aggregation indicate that parts of given quantities are 
to be considered as one. Thus, in 2 x (6 3), read "2 times the 
quantity (6 3)," the parenthesis requires that 3 be subtracted 
from 6 before multiplying by 2. 

Similarly, 12 + 4 (5 + 4) -s- (16 - 13) = 24. Explain why. 
Notice that according to rule 1 of this article 

3 a 2 means 3 ao, but (3 of means 3 a x 3 a, or 9 a 2 . 

Thus, 4 x 3 2 equals 36, while (4 x 3) 2 equals 144. 



8. Calculate (a -f &)c when a = 2, b = 3, c = 5. Ans. 25. 

Evaluate the following quantities when a = 4, 6 = 3, c = 2. 

9. 5a(6--c). 10. 5(a6 + c). 11. 3(a + 6c)a. 

12. 4(a 2 + 6 2 ). 13. 4(ac 2 -a 2 6). 14. 6(a6 2 -c)-2 a 2 6c. 

15. (4a6 2 ) 2 c. 16. 3(2ac) 2 6. 17. 



28. Exercise in Writing Algebraic Quantities. The student should 
fix in mind that if a, 6, and c are three numbers, 

the sum of a, 6, and c = a + 6 + c, 
while the product of a, &, and c = a&c. 
Also, that 3 fo'mes a = 3 a, while the third power of a = a 3 . 
Thus, if a is 5, then a 3 = 125, while 3 a = 15. 



30 THE ALGEBRAIC NOTATION 

Express the following in algebraic symbols : 

1. Six times a, plus three times b. Ans. 6 a + 3 b. 

2. Seven times a squared, less four times b cubed. 

Ans. 7a 2 --46 3 . 

3. Eight times x cubed, minus eleven times y cubed, plus three 
times y squared. 

4. Twice a cube, diminished by three times the square of a. 

5. The sum of a cubed, three times c squared, and the product 
of , 6, and c. 

6. The sum of a, b, c, and d. 

7. The sum of the squares of a and b. 

8. The double of x. 9. The difference between a and 5. 

10. The product of 3, x, y, and z. 

11. Four times the fifth power of m. 

ADDITION AND SUBTRACTION 

29. Addition of Literal Quantities. Solve the easier problems 
mentally. In the review solve a larger number mentally. 

1. Add 6 a, 7 a, and 3 a. 

SOLUTION. 6 a 

EXPLANATION. Here a represents some number, r.s 

o does every letter so used in algebra. The sum of 0, 7, 

and 3 times a number equals 16 times the number. 
10 a 

2. Add 4 a, 5 a, 9 a, and 6 a. Ans. 24 a. 

3. Add - 3 a, 6 a, - - 2 a, - - 4 a, 11 a. Ans. - 26 a. 

(See .6.) 

4. Add 4 m, 6 m, 9 m, 3 m. 5. Add - 5 b, - 6 b, - 7 b, - b. 
6. Add 24 a 2 , 15 a 2 , 13 a 2 . 7. Add - 11 a 3 , - 3 a 3 , - 2 a 3 . 



ADDITION 31 

8. Add 5x, Sx, 14 #, -11 a', and loo?. 

SOLUTION. 5x + Sx + 15 z = 28 x ; ( 14*) + ( llz) = 26x; 
28 x + ( 25 a:) = + 3 x. Ans. 

CHECK. 5x + 8 x = 13 x ; 13 x + ( 14 x) = x ; - x + ( llx) 

- 12 x ; - 12 x + 15 x = 3z, 
which is the same answer as before. 

a. REMARK. To make the problems of this article seem more real, think 
of the positive terms as denoting credits and the negative terms as denoting 
debts. Thus, the sum of $5x, $8z, and $15 a; is $28 a- ; the sum of one debt 
of $ 14 x and another of $ 11 x is a debt of $25x ; then the sum of a credit of 
$28 x and a debt of .$ 25 x is a credit of $3 x. 

Should a student be confused at any time in the process of adding alge- 
braic numbers, the difficulty will in all probability vanish if he will think of 
his numbers as denoting credits and debts. 

Add and check in the following problems as in Ex. 8. 

9. 7 x, 3 x, 11 x, 2x, - 5 x, - - 3 x. 
10. 2y,7y, 5y, - y, 10 y, --6y, Sy. 
11.5 x 2 , 12 x 2 , - 10 x 2 , x 2 , - 16 x 2 , 3 x 2 , 4 x 2 . 

12. 3 ax, - 2 ax, 5 ax, ax, 4 ax } - 9 ax. 

13. 2bc, --7bc, 3 be, 4 be, 5 be, - - 6 be, 5 be. 

14. -4 a 2 15. 7 be 16. 11 a' 2 b 2 17. 20 
+ 5 a 2 -86c -20-^ +15 
- 7 a 2 -96c +11 a 2 b 2 + 30 



18. 3a 2 6 19. 6a6c 20. 5 ax* 21. -25 a 3 ?/ 3 

4a 2 6 -21a6c -11 ax 3 -12arV 

a 2 & 3 abc ax 3 39 x^y 3 

-10 a 2 6 +12rf&c 7 aa 3 - 10 .T 3 ? 8 



22. What is the sum of 3 a 2 b and 5 air ? 

SOLUTION. Since the two given quantities are not similar ( 23), their co- 
efficients cannot be added for the coefficient of the sum. All that can be 
done is to indicate the addition. Thus, the sum is 3 a 2 b + 5 ab 2 . 

b. Indicating addition is called adding in algebra. 

23. What is the sum of 3 a, 5 b, and c ? 



32 THE ALGEBRAIC NOTATION 



24. Simplify 6 

SOLUTION. 6a + (- 3 a)+ (- a) = 2 a ; 
then the answer is 2 a -f 8 6 2 c. 

CHECK. Let a = 2, b = 3, c = 1. Then, given quantity, 
6 a + 5 6 3 a + 3 6 -- 2 c a = 
12 + 15 - G + 9 2 -2 =26; 
also, answer, 2a + 86 2c = 4 + 24--2=26. 

Since the answer 2a + 8& 2c has the same numerical value ( 24) as 
the given quantity, the presumption is that there has been no mistake in the 
solution. 

Simplify the following polynomials and check the answers to 
Ex. 25-29 as in the problem just considered (see 2, p. 6) : 

25. 2a + 2b + 2c a 3b + c + 3a. 

26. a b + 2c 2a + 3b 4 c. 

27. 2x + y 3x + z 3y+5x 

28. 7x lly + 4 : z 7z + llx 

29. 2 ay 3 ac 4 ay + 4 ac 6 ay -f- 5 ac -f- 11 a?/. 

30. 4 a? 2 3 a?;?/ + 5 y 2 + 10 xy 17 2/ 2 11 x 2 - 5 a??/. 

31. What kind of terms can be combined into a single term ? 
Ann. Similar terms (23). How are dissimilar terms added? 

See Ex. 22, p. 31. 

32. Make a rule for simplifying a polynomial ( 21) by answer- 
ing the following questions : 

(1) What is done with each set of similar terms ? 

(2) In adding a set of similar terms, how is the coefficient of the 

sum found ? (See 6.) 

(3) What is done with the results obtained ? (See Ex. 22, p. 31.) 

33. 2 xy 5 y 2 + x 2 y 2 1 xy + 3 y 1 4 o&y*- -\- 5 xy. 

34. 5 am 3 a?m 2 + 4 - - 4 am + 2 m 2 2 + T> + a L 'w 2 . 

35. Ila 2 &-7afr 2 + 2ac 2 + 10a&-4ac 2 + 5rr'6-4a& 2 . 

36. 3 x 3 - 2.2 or 2 + .4 ar 3 - 7.3 x + .5 or* - 1.5 .r- - .6. 

37. 240 a - 306 6 + 205 a - 70 a + 79 6 - 25 & + 100 c. 



ADDITION 33 

38. Add the following polynomials : 7 cr + 5 cm ; - 3 cm 4 d ; 
5 cm -+- 3 e 4 d 6 a 2 ; and 4 a, 2 - 7 cm d ; and check answer by 
letting a = 2, c = 3, d = 1, e = 4, m = 2. 

SOLUTION. Check. 

7 a 2 + 6 cm =28 + 30 = 58 

-3cw-4d -18-4 =-22 

_ 6 a 2 + 5 cm - 4d + 3 e = - 24 + 30 -- 4 + 12 = 14 

4 a 2 - 7 cm - _d _ = 16-42-1 =-27 

6 a 2 - 9 d + 3 e 23 

CHECK. =20 - 9 -f 12 = 23 

EXPLANATION. The first polynomial is set down, and then underneath it 
the other polynomials are written in such manner that similar terms ( 23) 
fall in the same columns. As fast as terms appear not similar to any above, 
they should be written in new columns at the right. Each column is then 
added as in Ex. 14-21, p. 31. 

To check the result the numerical value ( 24) of each polynomial is found 
and the results are added. If the sum found is the same as the numerical 
value of the answer, the answer is said to check. Thus, the sum 23 of the nu- 
merical values of the given quantities is the same as the numerical value of 
the answer. 

c. That a result checks does not prove that it is correct. Mistakes may 
have been made in both solution and checking. In rare instances results 
check and yet are wrong. Putting the letters equal to 1 or is less likely to 
uncover a mistake than giving them other values. 

Add, in the following problems, and check the answer in the 
first four, following the preceding solution as a model: 



39. 

40. 7a-46 + 2c;6a + 35-5c; -12a + 4c. 

41. x + y + z-, x~2y + 2>z; -5x y + z + 2u. 

42. 3a6 + 5ac; 6a& Sac; Sab ac + abc. 

43. 5x 3a + 6 + 7; 4 a 3 a + 26 9 ; x b. 

44. 3 a _5_|_2c; 5a + c 25; 36 a--4c. 

45. 20p q+r; 2p + 5q~7r; - 
46. 



34 THE ALGEBRAIC NOTATION 

47. 2ab-3ax 2 + 2a 2 x; 12 ab - 6 o?x + 10 ax 2 ; ax*-8ab-5a?x. 

48. 3 a? 2 - - xy + 072; - - 3 y 2 + 4 yz z 2 ; Stfxy xz + Syz-, 

6a?-6y-6z; 4=yz - 5yz + 3z 2 -, - Ix 2 + y 2 + 3yz + 3z*. 

49. Make a rule for the addition of polynomials telling : 

(1) How the different polynomials are set down for the operation ; 

also how similar terms are placed, and where dissimilar 
terms are written as fast as they appear. 

(2) What is done with each column. How it is done. 

(3) How the answer is checked for correctness. 

50. 3a 2 b s -7 ab 4 +5axy; -7a 2 b 3 -2ab 4 -axy- ab*-7 axy+8a 2 b 3 ', 

- 10 ab* + a 2 b 3 + 3axy; - 5 a 2 b 3 + 18 

51. 2ab + 3ax 2 +2d 2 x 2 : > 22 ab-6a 2 x 2 +Wa 

52. 10 a 2 b - 12 a 8 6c - 15 b'c 4 + 10 ; 8 a?bc - 10 b 2 c* - 4 a 2 b - 4 ; 

20 6V - 3 a s bc - 3 a~b - 3 ; 2a 2 b + 12 a s bc + 5 6 2 c 4 + 2. 

53. Jaj 2 -2a 8 + |a;-3; i^ 3 - f x 2 + J; 2aj-|x 2 . 

54. Of two farmers the first had 2 a? 3 ?/ acres, and the second 
had x--y acres more than the first. How many acres had the 
second ? How many had both ? 

30. Subtraction of Literal Quantities. We saw in 7 that sub- 
traction is the process of finding a number which added to one of 
two given numbers, the subtrahend, produces the other, the minu- 
end. Solve the easier problems mentally. 

1. What is the difference between 6 a and 4 a ? Ans. 2 a. 

2. What is the difference between 11 a 2 and 3 a 2 ? 

3. 7a 3 -2a 3 =? 4. 20 abc - 14 abc = ? 5. 15 or 5 ?/ - 11 y?y = ? 
6. What is the difference between 11 m and 4m? 

SOLUTION. llm EXPLANATION. 15?w + ( 4 m) = 11 m. 

-4m CHECK. A credit of $15w and a debt of 
15 m amount to a credit of $ 11 m. 



SUBTRACTION 35 

7. What is the difference between 12 x and 10 ./: ? (See 7, a.) 

8. What is the difference between 4 x and 6 x ? Ans. - - x. 

9. Ux-(+16x)=? 10. -llx-(-ox) = ? 

In the following subtract the lower quantity from the upper. 
Check the answer by thinking of the quantities as debts and 
credits : 

11. -15 a 2 12. -21 a 3 13. -9afcc 14. 4crb 

5 a 2 - 5 a :i 13 afrc 3 a'b 

15. llm 2 16. -15 a 2 17. -6 a 3 18. 20 x 2 
-12m 2 -25 a 2 +15 a 3 - ( .U 2 



19. 12 y? 20. -17?/ 2 21. 22. 

18 a 2 -lly* 12 xy -15??m 



23. Make a rule for subtracting one monomial quantity fro-.n 
another similar quantity. 

(1) In what two ways can the problem be set down ? (See Ex. 9 

and 11.) 

(2) What quantity must the difference be ? ( 7, 23.) 

(3) How can the answer be checked ? 

24. -12abc 25. -6a 2 6 26. 19 ab 2 27. 7 37/2 
+ 18 abc - 15 a 2 b -7ab 2 - 11 xyz 

28. From the polynomial c~x 2 + 3 ex 2 - 5 ex --4 cc 2 take (?3?--2cx 
-f 3 c 2 ^* 6 c 8 , and check the answer with c = 2, it* = 1. 

SOLUTION, c-x 2 + 8 ca: 2 - 5 c.r -- 4 z 2 4 

c%2 _ -2cx _ + 8 c-x - 6 c 3 - - 36 

3 rx -2 _ 3 cx _ 4 x -2 _ 3 c % + 6 c 3 =: + 32. Check. 

EXPLANATION. Terms of the subtrahend similar to those of the minuend 
are placed in the same column ; dissimilar terms are put in new columns. In 
the second, fourth, fifth, and sixth columns zeros are supposed to take the 
place of the blank spaces. Notice that in each column, the remainder is chosen 
so that the subtrahend and remainder added algebraically give the minuend. 



36 THE ALGEBRAIC NOTATION 

29. From 6 a 2b 5c take 2 a 2 b 3 c. Check. 

30. From 7 x 2 8 x - - 1 take 5 x 2 6 cc + 3. Check. 

31. From 2 x 2 - 3 a 2 ^ 2 + 9 take or + 5 a 2 x 2 - 3. Check. 

32. From X s 3 3?y + 3 x*y 2 take - - x 2 y -f 5 a,* 2 ?/ 2 y*. Clieck. 

33. (or - 5 xy - y 2 + 3 yz -2 z 2 ) - (x 2 +6 #?/-fll x*z-4 y 2 -2z 2 ). 
34. 

35. 

36. (a 2 b* - 6 a 2 c + 9 ac 2 ) - ( - 2 crc 2 - 6 acx + 3 a 2 ^). 

37. 

38. 



Simplify the following expressions (as in 29, Ex. 24-37) and 
then take the second from the first : 

39. 4 #y 5 cz -{- 8 m - - 4 cz 2 m, and cz -f 2 x 2 y* 4 cz. Check. 

40. 4 a? 3 3 x 2 -- 2 x Atfx+l, and 7 x 3 x 2 x 2 Sx 3 11 + 3 x. 
*41. 3 3 -2a 2 -7a 3 -5a-2a 2 , and a 3 -4a 2 -5a-lla 2 -14. 

42. 2 m 3 3 m 2 n -f- 6 m 2 n + ?i 3 , and m 3 + 12 mn 2 - 12 m 3 . 

43. From a 3 - '6 a 2 c + 9 ac 2 take the sum of - 2 2 c - 4 a 3 -f 2 ac 2 

and - 2 a 2 c + 3 a 3 ac 2 . 

44. From the sum of 36 + 2a--4c and 9c + 36--5d subtract 
the sum of 6 d - - 1 a and 8a 7^ + 96 + 5 c. 

45. From the sum of 7 X s 4 a? 2 -f 6 # and 3 x 2 - - 10 x 5 take the 
sum of 5 3 + 4x + 12 and 8 x 3 11 o^ 2. 

46. A man who had four sons gave to the youngest $a, to the 
next older $ b more than to the youngest, to the third son c + 50 
dollars more than to the second, and to the oldest as much as he 
gave to the three younger sons. How much more did the oldest 
get than the youngest ? 

47. One man has 2 a b + c dollars, and his brother has 
a 2b + d dollars. What is the difference in their fortunes? 

* See 1, p. 6. 



SUBTRACTION 



37 



48. Lines can be denoted by letters by letting each letter rep 
resent the number of units of length in its line. 
(1) Add -3 and +7. 

SOLUTION. We first 
measure out 3 units to the 
left because 3 is negative 
(page 16). From this point 



-3 



0-3 +0-T) = + 4Ans. 

we measure 7 units to the right, arriving at -f 4. Thus, 3 + 7 = + 4. 

(2) Add -5 and +12; 
> 1 also +5 and +8; also 

-f- 9 and - 6. 

(3) Show that 



-1 



+ lAns. ' 

See diagram. 

(4) Make problems similar to that in last diagram and solve 
them. Check, solving as in 9. 

31. Problems. (Continued from 1.) 

1. Find what number x is when 3 x -\- 1 = 13. 

SOLUTION. If 3# increased by 1 equals 13, 3 x must equal 1 less than 13, 
or 12. If 3 x = 12, then x = | of 12, or 4. Ans. 

PROOF. 3 x 4 + 1 = 13. (See 27, 2.) 

Solve and prove the following in the same way : 

5. 7o; + 2 = 44. 6. 10x + 5 = 55. 7. 7oj--4 = 3 

SOLUTION to Ex. 7. If 7 x diminished by 4 equals 31, Ix must equal 4 
more than 31, or 35. If 7 x == 35, then x = \ of 35, or 5. Ans. 

PROOF. 7x5-4=31. 
8. lloj-3 = 30. 9. 10a?-17 = 33. 10. 15a-23 = 

SUGGESTION to Ex. 13. The difference between 9x and 4x is 15. 

14. 12x 72 = Sx. 15. Sx 14 = #. 16. 14# 24 = * 

17. 10# = 5o; + 50. .18. x = 6x 2. 19. 3x 12 = a?, 



38 THE ALGEBRAIC NOTATION 

20. The difference between two numbers is 6 and their sum is 
76. What are the numbers ? 

SUGGESTION. If x = smaller number, what expression with x in it equals 
the larger ? Am. x +6. Then x + (x + 6) equals what number ? Now 
solve 2 x + 6 = 76 as in the preceding solutions. 

21. The difference between two numbers is 11 and their sum is 
97. What are the numbers ? 

22. Two men earn $180 a month, but one earns $10 a month 
more than the other. How much does each earn ? 

23. The sum of the ages of three girls is 37 years. The first is 
3 years older than the second, and the third two years younger 
than the second. What is the age of each ? 

SUGGESTION. Let x = number of years in age of second. Then what 
is age of oldest ? Of youngest ? Adding these ages, we have 

(x + 3) + x + (x - 2) = 37 
or, 3x + 1=37. 

24. The sum of the ages of three men is 157 years. The oldest 
is 6 years older than the second, and the second 5 years older than 
the third. What is the age of each ? 

25. A farmer has a doubletree 4 ft. 4 in. long which he wishes 
to divide so that the longer part will be 1 in, longer than twice as 
long as the shorter part. How long shall he make the shorter 
part ? Prove answer. 

26. Divide $ 500 among A, B, and C so that B shall have $20 
more than A, and C $55 more than B. Prove. 

27. Three times a given number diminished by 20 is equal to 
the given number. Find the number. . See Sug. under Ex. 13. 

28. At an election two opposing candidates received together 
2150 votes and one received 98 votes more than the other. How 
many did each receive ? Prove. 

29. Three men A, B, and C, keep their cows in the same pasture 
and together pay $ 56 for the use of it. A has one cow, B has 3, 
and C has as many as both A and B. How many dollars will 
each have to pay ? 



PROBLEMS 



39 



30. A boy bought an equal number of apples, lemons, and 
oranges, paying 1^ for an apple, 2/ for a lemon, and 3^ for an 
orange. How many of each did he buy for 72 ^ ? Prove. 

SUGGESTION. Let x number of each sort. Then how many cents did 
the apples cost ? The lemons ? The oranges ? 

31. A horse, saddle, and bridle together cost $ 144. The saddle 
cost 11 times as much as the bridle, and the horse 15 times 
as much as the saddle and bridle together. What was the cost of 
each ? 

a. REMARK. It is important that the word number always come after 
" Let x = ." Beginners often write : Let x = the bridle, or x = the cow, or 
x = A, or the like. Such statements are not clear, and are pretty sure sooner 
or later to lead to confusion of ideas. 

32. A gate 49 inches high is made 
of four boards each 8 in. wide. The 
second space is to be 21 in. wider 

*^ 

than the lowest, and the upper 3 in. 
still wider. Find the width of each 
space. 

33. Frederick has his money in dimes, nickels, and cents, and 
he has the same number of each kind of coin. The value of all 
is 80^. How many pieces of each kind has he? 

34. The leaning tower of Pisa in Italy is 179 ft. high, which 
is 15 ft. more than 12 times the distance it leans over. How 
many feet does it lean over ? 

35. In midwinter at St. Petersburg the night is 13 hr. longer 
than the day. What is the length of each ? 

36. A rectangular field is twice as long as it is wide, and the 
distance around it is 372 yards. What are its dimensions ? 

SUGGESTION. Draw a figure and write x beside each short side. Then 
what shall be written beside each long side? Express the whole distance 
around by adding the lengths of the sides. 

37. At an election there were two candidates, and 1280 votes 
were cast. The successful candidate had a majority of 40. How 
many votes were cast for each candidate? 




40 THE ALGEBRAIC NOTATION 

38. If the total wealth of the United States in 1900 was 88 
billions of dollars, and the total wealth exclusive of money in cir- 
culation was 43 times all the money in circulation, what was the 
number of dollars of money in circulation ? 

39. If there were approximately 4 persons not in public school 
to every enrolled pupil, and 36 pupils to every teacher, how many 
public school teachers, pupils, and other persons were there in the 
United States in 1900 when the population was 76,201,000 (using 
this instead of the census number, 76,303,387) ? 

40. The total population of the United States in 1900 was 
76,303,387 and there were 1,815,097 more males than females; 
how many were males and how many females ? 

41. The sum of the exports and imports of the United States 
during the year ending June 30, '07 was $ 3,315,272,503 and the 
exports exceeded the imports by $ 446,429,653. What were the 
imports and exports ? 

42. The mill consumption of cotton by the world in 1906-1907 
was 19,493,441 bales of an average weight of 500 Ib. Of this 
amount the United States raised 928,913 bales more than twice 
the product of all other countries combined. How many bales 
did the United States produce ? 

SYMBOLS OF AGGREGATION 

32. Removal of Symbols of Aggregation. 

1. Simplify (3 x + 2 y) + (4 x - 3 y). 

SOLUTION. The sum is the same whether the quantities are tied together 
in groups by parentheses, or are left untied. 

Thus, (3o; + 2y) + (4a;-3y) = 

3 x + 2 ?/ + 4 x By =7 x y. Am. 

CHECK. Let x 2, y = 1. 

Then, the given quantity, (3 x + 2 ?/) + (4 x 3 y') 8 + 5 = 13 ; 
also, the answer, 7 x y = 14 1 = 13. 



SYMBOLS OF AGGREGATION 41 

From this solution and the explanation preceding it, we can see 
the truth of the following general rule: 

2. Symbols of aggregation preceded by +-, or by no sign, can be 
removed without changing the sign of any term inside. 

a. It must be remembered here that + is understood before the first 
quantity within a symbol of aggregation, if no sign is written ( 8). 

3. Simplify (3 x 4 y) + (2 x + 7 ?/), and check answer. 

4. (3x 2 2x) + (2x-- 1). 5. (b 2c)+-(4c 56). 

8. m 2 -3m + S5m-2w 2 |. 9. (6 a; + 2 y) + [4 a; - 11 y]. 

1 f\ } j'\ -t / r\ r\ 

J. U t/ "" r t/ O "T~ tl C/ 



SUGGESTION. After solving by removing the vinculums, rewrite the 
problem, using parentheses, and then solve. 

11. Simplify (3x + 2y) (a-- b), and check. 

SOLUTION. This means that a b is to be subtracted from Bx-\-2y. 
Writing the quantities down for subtraction as in 30, Ex. 28, and seeking a 
quantity which added to the subtrahend equals the minuend, we have 

2y 

a b 



3x + 2y - a + b 

Or, 3x + 2?/ - (+ a- &) = 

3x + 2y a -}- b. Ans. 

CHECK. Let x = 4, y = 4, a = 3, b = 1, Then 
given quantity = (3 x + 2 y} (a - b} = 20 2 = 18 ; 
also, answer =3x + 2y a + 6 = 12+8 3 + 1 =18. 

From this solution we can see the truth of the following general 
rule : 

12. Symbols of aggregation preceded by can be removed by 
changing the sign of every term inside. 

b. Notice that the + or sign before a symbol of aggregation drops out 
when the symbol is removed. It denotes an operation, and this operation is 
performed when the symbol is removed as explained in 2 and 12 above. 



42 THE ALGEBRAIC NOTATION 

Simplify the following. Check the answers to the first seven 
and any other when in doubt about the correctness of the result. 

13. (3&4-2c)-(& 4c). Ans. 26 + 6c. 

14. (4m-2?i)-(2w 5w). 15. (4 a 2 - & 2 )-(2a 2 + 2 fr 2 ). 
16 . (x 2 + 2 xy) ~(x 2 2xy). 17. a& - - (m -- 3 aft + 2 ax). 
18. 9m ( 4 ??i + Gn n 2 ). 19. -[a 6] [6 ~ c]. 

20. 



c. We bad just such problems as the preceding ones in 30, commencing 
with Ex. 33. There the process of solution was different. To show that re- 
moving the parentheses and then simplifying, as we have just been doing, gives 
the same result as writing the subtrahend under the minuend and subtracting, 
we will give the two solutions. Notice that the answers are exactly the same, 

Solution by removing parentheses and simplifying (adding). 

(3 a; 2 + 2 xy + y*)- (z 2 -2xy + ?/ 2 ) = 

3 x 2 + 2 xy + y' 2 x* + 2 xy - y 2 = 2 x 2 + 4 xy. Ans. 

Solution by the process of subtraction. 



2 x 2 + 4 xy. Ans. 

In this article the problems are to be solved, of course, by the process of 
removing symbols. 



21. (4a + &) (x + 4:a+b 2y x y). 

22. 4?/ 2 - 

23. 3a 2 - 

24. (5a-46 +3c)-(-3a + 26 - c)-(- a - c). 

25. Make a rule for removing symbols of aggregation : (1) What, 
is done when the symbol is preceded by -f- ? (2) When preceded 
by - - ? What is done after the symbols are removed ? (See 29, 32.) 

26. x 3 + x>- (x - 






27. a?- 



SYMBOLS OF AGGREGATION 43 



28. a 6 2<H 2 a + b +3c a 2 b + c. Ans. 2a + 2b. 

d. The vinculum over a quantity means the same thing as parentheses 
around it. In the first and third quantities the sign + is understood before a. 
See a, p. 41. 



29. 3 m + 2n 3 m 2n -+- 9 m -- n. 



30. 3a 6 + 7c--2a + 36 56 4c + 3c a. 

Symbols of aggregation oftentimes appear inside other symbols 
of aggregation. 

31. ab c {oj + 5 (c ab + 2x)\. 

SOLUTION. abc{x + 6 c-}-ab 2x} (On removing the parenthesis.) 
ab c x 5 + c ab + 2x. (On removing the brace.) 

x 5. Ans. (On simplifying as in 29, 32.) 

CHECK. Let a = 6, b = 2, c = 2, x = 4. Then 

(jwew quantity ab c {x + 5 (c ab + 2 #)} 

becomes 12 - 2 -{4 + 5 - (- 2)}= 10 - { + 11} = -1 ; 
also, answer, x 5 = 1. 



32. 2a-2&- 

33. 3a + x [a + 5x+ (3 a 2 a;)]. 
34 . a [6 + (c + 2 #) ( y .r) ] . 

35. 1 a?-- 51 --x [I'-aj + aj 2 ] 3x\. Check. 

36. 2a 



37. Make a rule for removing symbols of aggregation when one 
set of symbols is within another set. Which are removed first ? 
Ans. The inside ones. After the symbols are all removed, what 
remains to be done ? How is the work checked ? 

38. mx- -[Sy + (6a mx) 2 ]. 

39. 3x 

40. a b a b~ 

41. tt- 



44 THE ALGEBRAIC NOTATION 



42. l.-2-l-x-x*-x. 



SUGGESTION. First, remove vinculum ; next, remove parenthesis ; third, 
remove brace ; fourth, simplify. 



43. 2a-[a-j&-(36 2a)j]-(&-a 

44. 8a-2b 



33. Inserting Quantities within Symbols of Aggregation. 

1. Any terms can be inserted within symbols of aggregation pre- 
ceded by. + without altering their signs, since this amounts merely 
to grouping them in a certain way. See 13, 2. 

2. Any terms can be inserted within symbols of aggregation pre- 
ceded by by changing the sign before every term so inserted 
See 32, 12. 

In the following, check the answers by removing symbols again, 
noting if result is the same as the given quantity : 

3. Change a + b 4- c into a binomial by inclosing the first two 
terms within parenthesis. 

4. Change 2 + 2a& -f b 2 -c 2 - 2cd-d 2 so that the first three 
terms appear in one parenthesis preceded by +, and the last three 
terms in another parenthesis preceded by -. 

5. Make a binomial out of a b -f- c d. 

Q. Insert the first two and last two terms of 3x 2 2xy 6xz 
4- 4 yz in two parentheses, making precede the second. 

7 Change 2 2 ab + b 2 2 ac c 2 -f 2bc--a + b c into a tri- 
nomial, putting a 2 , b' 2 , and c 2 in the first parenthesis, --2 ab, - - 2 ac, 
2 be in the second, and - - a, b, and - - c in the third, the second and 
third parentheses preceded by . 

8. Put the terms of a b -f c d --ef+g h in parentheses 
two by two, (1) beginning with the first term ; (2) beginning 
with the second term. Choose the sign outside so that - - does 
not precede the first term inside in each instance. 



MULTIPLICATION 45 

MULTIPLICATION 
34. Exponents in Multiplication. Read 16 carefully. 



1. Multiply 3 by a 2 . ^Uliat does a 3 mean? a 2 ? What is 
the product of aaa and aa ? Ans. aaaaa, since no sign be- 
tween factors in algebra denotes multiplication. How is aaaaa 
written in the exponent shorthand? Ans. a 5 . Thus, we have 
a 3 x a 2 = a 5 . 

How can the exponent 5 in the product be found from the ex- 
ponents of the factors ? Ans. By adding the exponents of the 
factors. 

2. a'-xa 2 . 3. a 4 x a 3 . 4. a 5 x a 4 . 5. 6 4 x6 2 . 

6. b x b\ Ans. 6 4 . Why ? 7. a 3 x a 2 X a 2 . 8. c 5 X c 7 X c. 

9. W 2 xm 6 . 10. w 5 Xm 5 . 11. m 3 X m 2 X m. 12. w 4 X n 3 x >r. 

13. Make a carefully worded rule governing the exponents in 
multiplying when the same letter appears with exponents in two 
or more factors. 

14. a 7 x a 2 . 15. s? x or 5 X a 2 . 16. a; 9 x a 5 X x. 17. a 2 X a 2 X a 2 . 

35. Multiplication of Literal Quantities. Find the product of 
monomials mentally. 

1. What is the product of 4 a?b and 5 abc ? Ans. 20 a s 6 2 c. 

EXPLANATION. The sign of the product is because unlike signs give 
( 10, 5) ; the coefficients 4 and 5 are multiplied together for the coefficient 
of the product ; a~b x abc = (a 2 x a) x (b x &) x c by the order and group- 
ing laws in multiplication ( 13, 3, 4) ; then, a 2 x a = a 3 , b x b = 6 2 , as in 
the preceding article. 



2. 3a-x2ab. 3. 4a 3 6x -3a 2 c. 4. 12 x 2 y x 2 ajy. 

5. 3 ac x 5 b. 6. 5 a 2 # x - - 7 ax 3 ?/. 7. ,4 aw x 3 am 3 X am. 



46 THE ALGEBRAIC NOTATION 

8. - a 4 z x - a s x 2 z. 9. 4 ale 3 x -- 3 ax 2 . 10. 5 Irx* x - - 7 

11. 2m 3 ce 2 12. -8a/n/ 2 13. f a 6 raa; 4 

x - - 3 m# 4 X - - 5 a?/ 3 x --|fom 3 a? 



14. Make a rule for the multiplication of literal quantities, 
answering the following questions : 

(1) What sign will the product have if there is an odd number of 

negative factors ? If there is not an odd number of nega- 
tive factors? See 10, 25. 

(2) How is the coefficient of the product found ? 

(3) What is done with the exponents of a quantity that appears 

in more than one factor to get the exponent of such quan- 
tity in the product ? What is done with literal quantities 
that appear in only one factor ? 

15. 4 ax 2 x - - 3 aV. 16.2 mn 2 x -- 3 m 2 n X - - 2 m 3 w x - - 4 mn. 
17. - 5 x"y x - - 4 xyz. 18. -5x--3ax--26x--cx4a. 

19. - 11 arty X 4 arty 2 z. 20. - G m 2 x -- 3 m 2 x -- m 3 x - - 4 m 4 . 
21. X G a-'V;. 22. 3 &c x -- 2 a 2 fcc X ~> /*: 2 X 0. 



a. By " taking " we can understand removing an object from one place to 
another. Then taking any quantity no times evidently gives nothing; like- 
wise, taking nothing any number of times gives nothing. 

23. 4a 2 or 3 24. xifz 1 25. forty 26. -5an s y 2 

X 3 ax 2 X x 3 ?/ 2 x - - 4- xy 3 X - - 4 a 4 hx 

27. -3mn 2 28. 1.3a 8 6 29. 1.7 abc 30. -fa s m 2 
X - - 4 mn 4 x .5a6 3 - ,2abd X - - 



n 



31. Multiply m by ?,, the letters denoting number of units of 

length in lines. Suppose m and /i denote 
the lines underneath them. Then the 

n 

number of units of area in the rectangle 
formed with m line as length and n line 
as width equals the product of m and n, 



inn 



or mn. 



MULTIPLICATION 47 

32. Multiply a + b by c, when each letter is supposed to repre- 
sent the length of a line. 



a a b 


a + b 
e x c 


b c 


ac 


be 





ac + be 
33. Multiply 2 ax 2 3cz by 5 az and check with a = 2, c = 2, 

SOLUTION. 2 ax 2 3 cz = 14 

x 5 az = x 30 



10a 2 z 2 2 -- 15acs 2 = --420 

34. Multiply 8 cr 4 a# by 4 am 3 and check. 

35. (4 a 2 - 3 6) x 3 ab. 36. (8 a 2 - 9 ab) x 3 a 2 . 
37. 2a(a 2 6 2 -2a&). 38. 2 afy(3 a 2 - 4 / + 5 z 2 ) 
39. -5oj 2 2/(3^-2a- 2 z/+2/ 3 ). 40. 2 a&ceZ(a + 6 + c + d). 

41. 4 a 2 3ac + 2 42. ??i# + ?iy - - 10 z 

X 5 ax X wi 2 n 



43. Make a rule for finding the product of a polynomial and 
monomial, telling what terms of the polynomial are multiplied by 
the monomial, and what rule applies when any two terms are mul- 
tiplied together. 

44. In these problems the distributive law applies. What is it ? 

45. -9a6(5a-2a& + 10). 46. (-9 a 5 +3 a*b*-W) X -Sab*. 



47. Sx 2y --4 + 5 48. f-aV-- f a^--| x 

Xx x a?x 



49. The student must note carefully the difference in the man- 
ner of forming the product of two monomials and of a polynomial 
and a monomial. 

Thus, 3 x abc = 3 6c, and does not equal 3ax3Bx3c = 27 abc. 

But 3x(a + &+c)=3a+36+3c, as in the problems just solved. 



48 THE ALGEBRAIC NOTATION 

50. To multiply the product of several factors by a given quantity, 
multiply one of the factors by this quantity, leaving the other factors 
unchanged. 

51. To multiply the sum of several factors by a given quantity, 
multiply each term by the given multiplier, and add the results. 

52. 6x2abc. 53. 6 X (a 2 + b 2 + c 2 ). 
54. 2 x3 X 4(m + ft+p). 55. 2 m x 3pqr X 4n 2 . 
56. 6x7x5 xy. 57. 6(7 + 5 xy). 

58. Multiply a + b by c + d and check with a = 2, b = 3, c = 1, 

c? = 5. 

DIAGRAM ILLUSTRATION 

SOLUTION. 



a - 
c - 


-b 

\-d 


^bd 


- 5 a + & 
x 6 




ac 


be 


ac - 


h be 
+ ad- 


30 c 


ad 


bd 


ac A 


- be 4- ad -i 


\-bd 


-30 + ^ 



EXPLANATION. Notice that each term of the multiplicand is multiplied by 
every term of the multiplier. (See 13, 5.) 

59. Multiply m + n by p + q and check with m = 4, n = 2, p = 5, 
g = 3. 

60. (a? + 5) (a; 4- 10). Check with # = 4. 

SOLUTION, x + 5 =9 

a; + 10 = 14 

Z 2 + 5x 126 

+ 10 x + 50 



x 2 + 15 x + 50 = 126 

61. (aj + 8)(a? + 8). 62. (x- 7)0 - 10). 

63. (a; -12)(a5-l). 64. (2x-3)(x + 5). 

65. (a? -15) (-a; + 3). 66. (a; +19) (a; -20). 

67. (2 a? + 3) (a? -7). 68. (a? + 3 /) (a? + 4 y). 

69. (a --9 6) (a + 5 6). 70. (xy ab)(xy + ab). 

71. 2-3r2 + 3r. 72. (- 



MULTIPLICATION 49 

73. Multiply 3 m~ + mn 2 n 2 by m 2 - 6 n 2 , and check with 
= 3, n = 2. 

SOLUTION. 3 m 2 -f mn 2 w 2 = 25 

m 2 - 6 n z = x -- 15 

3 w 4 + m*n 2 m' 2 w 2 - 375 

- 18 i2 w -2 - 6 mn* + 12 n* 



3 w 4 + w 3 - 20 m 2 n 2 - 6 mn* + 12 n 4 = 375 

74. (a 2 - 2 (wj + 4 a? 2 )(a 2 + 2 ax). Check with a = 3,x = 2. 

75. (4a 2 + 6.ivy-h92/ 2 )(2a--32/). Check with x = 3, y = 1. 

76. (x> - xy + f)(x + y + 1). (See 13, 5.) 

77. (a 2 - 

78. (_a 5 

79. (x 2 3 xy y 2 )( xr -\-xy 

80. (2^-3ar + 2x)(2^ + 3 

81. Multiply m + ?i +p by r + s + t and explain the result with 
a graphical illustration as in Ex. 58. 

82. Make a rule for multiplying one polynomial by another, 
telling 

(1) How to set the problem down. 

(2) What monomial multiplications are performed. 

(3) What operation is performed last. 

(4) How the correctness of the result can be checked. 

83. (l + 2a;-3a,' 2 + aj 4 )(^-2a;-2). Check with x = 2. 

84. (ct 3 -3a 2 64-3a6 2 -6 3 )(a 2 -2a6 + 6 2 ). Put a = 3, 6 = 1. 
85. 

86. 
87. 

88. (aj-2)(a?-3)(aj-4)(- 

89. (a 2 - b 2 ) (a 2 + ab + b 2 ) (a 2 - a& + b 2 ). 

90. (a 3 + 2 a 2 b + 2 a& 2 -f & 3 )(a 3 - 2 a 2 6 -f 2 a& 2 - 6 3 ), 



50 THE ALGEBRAIC NOTATION 

36. Powers and Roots of Monomials. (See definitions 17 and 18.) 

1. Square 3 a 2 . SOLUTION. 3 a 2 x 3 a 2 = 9 a 4 . Ans. 
2. (5a 3 ) 2 =? Ans. 25 a 6 . 3. (mV) 2 . 4. (-4a 3 6) 2 . 5. (-o?bcf. 

6. (-2 a 4 ) 2 . 7. (-i<> 2 . 8. (20 a 2 ) 2 . 9. (-^m 3 ) 2 . 10. (^6 6 ) 2 . 

11. Make a rule for squaring a monomial. 

(1) What is done with the given coefficient to get the coefficient of 

the answer ? 

(2) By what is each exponent multiplied to get the exponent of 

its quantity in the power ? 

(3) What sign does the answer always have ? 

12. Cube -4a 2 5. 

SOLUTION. -4 a 2 b x -- 4 a 2 & x 4 a^b = - 64 6 & 3 . 

13. (or 5 ) 3 . 14. (-z 2 ?/ 3 ) 3 . 15. (-2a 2 c 4 ) 3 . 16. (-6?/) 3 . 

17. (z 4 /) 3 . 18. (Jm 2 w) 8 - 19. (-5m 4 ) 3 . 20. (-3 a; 4 /) 3 . 

21. (-a 10 ) 3 . 22. (a-?/z 2 ) 3 . 23. (-6mV) 3 . 24. (4a^z 2 ) 3 . 

25. Make a rule for cubing a monomial. 

(1) What is done with the given coefficient to get the coefficient 

of the power ? 

(2) By what is each exponent multiplied to get the exponent of 

its quantity in the power ? 

(3) If the given quantity is positive, what sign will the power 

have ? If negative, what sign will the power have ? 

26. Make corresponding rules for higher powers. 

27. (-3 3 ) 3 . 28. (2 a 2 ) 4 . 29. (-3 a 3 ) 5 . 30. (-5aW) 3 . 

31. (-3 a; 4 ) 6 - 32. (-15 z 4 ) 2 . 33. (-6 a 4 ) 3 . 34. (-abc) 7 . 

35. What is the sign of an odd power of a negative quantity ? 
Of an even power ? What is the sign of all powers of positive 
quantities ? 

36. V36a 2 = ? Ans. 6 a. 

a. The sign is used to denote either -f or . Thus, (-f 6 a) 2 = 36 a 2 ; 
and, also, (-6 a) 2 = 36 a 2 . 



POWERS AND ROOTS 51 



37. VU5 iii 2 . 38. Vl6m 4 . 39. V8a*. Ans. + 2 a. 40. 



41. V64 a. 42. V<49a 8 . 43. ^G4~c?'. 44. </16e?. 45. J/rJ5~a* 
46. V ^ 47. V25 w 4 . 48. V32i 49. -\/~ - 27 a 3 

50. V225 a G . 51. A/243 a 5 . 52. V ; - a ;! 6 (i . 53. -v/aTT 8 . 



54. The same quantity can often be regarded as one power of 
one quantity and another power of another quantity. 

Thus, a 6 = (a 2 ) 3 = (a 3 ) 2 ; a 8 = (a 2 ) 4 = (a 4 ) 2 j 

x 12 = ( 2 ) 6 = (or) 4 = (x- 4 ) 3 - (a 6 ) 2 . 



It is important that the student understand and remember this 
truth. It will later be put to frequent use. 

37. Arrangement of the Terms of a Polynomial according to the 
Powers of a Leading Letter, and Arrangement of the Factors of a 
Single Term. 

In most polynomial quantities in algebra there is a leading letter. 
A polynomial is said to be arranged with reference to the powers 
of the leading letter when the exponents of this letter either in- 
crease or decrease in regular order. 

For example, 2 a 4 + 2 a ?J b + 5 a 2 6 2 --6 a& 3 -f- 4 b 4 is arranged with 
reference to the descending powers of a. On the other hand, 
l + 2# Go^ + Sx 5 is arranged according to the ascending powers 
of x. 

The letters of a single term are said to be arranged when they 
come in the usual alphabetical order. Thus, 6a*b 2 c 4 xy 2 . If the 
student will examine the various quantities given in multiplica- 
tion, he will find that they are nearly always arranged in the 
manner described in this article. In multiplication and division, 
and elsewhere, if quantities are given in a disarranged form, they 
should be immediately arranged. 



52 THE ALGEBRAIC NOTATION 

DIVISION 

38. Division of Literal Quantities. Perform division of mono- 
mials by monomials, mentally. 

1. Divide 24 a We? by -3a6 2 c 2 . Ans. -Sa 2 b 3 d. 

EXPLANATION. Why is the answer ? (See 11.) How many times 
is c 2 contained in c 2 ? How is the exponent 2 of a found ? How is the ex- 
ponent 3 of b found ? How is the exponent of a literal quantity in the 
quotient found from the exponents of the same quantity in the dividend and 
divisor ? 

2. 4 a 5 -2 a 2 . 3. 30 a*b 2 -=- 5 a 2 b. 

4. - 32 xyz -r- - 8 xy. 5. 42 cVn -r- - 3 c. 

6. 45 a% V -r- - 9 a s bx 2 . 7. -21 a 2 & 3 -r- -7 a 2 6 2 . 

8. -i^s^.j^s^ 9> 225 m?ij -r- - 25 my. 



30 ajy -Slabdif m*pW 42 

10. ' 1 J. . -- J./a. - * J.O. 



. . . . 

- 5 xy 2 3 My mpsr 7 XZ A 

14. Make a rule for division of monomials, telling 

(1) How the sign of the quotient is determined. 

(2) How the coefficient of the quotient is found. 

(3) How the exponents of the several literal factors are found. 

(4) How the solution is proved correct. 



15. - 16 x*y 2 + -xy\ 16. - 3 a 5 c 9 -h - 

17. 16b 2 x 2 y + -2xy. 18. 63 aW -- 9 aW. 

19. (a + b) 5 H- (a + b) 2 . Ans. (a + b)\ 

20. 4(m-|-7i) 4 -5-2(w + w) 8 . 21. 8 (a - &)V-=-2(a 
22. -50ary-*--5aty. 23. - 35 a 11 -- - 7 a 7 . 
24. - 1 x 4 ?/ 7 ^- 1 .T?/ 5 . 25. | m 8 n 7 -. f 



26. Divide 9 x 3 - - 12 x 2 y + 3 a by 3 x. 

SOLUTION. 3 x)0 r 3 - - 12 .r' : ji/ + 8 

3 x' 2 4 xy -f 



DIVISION 53 

PROOF. 3 x z 4 xy + z 



- 12 x' 2 y + 3xz 

a. Notice that when a polynomial is multiplied by a monomial every term 
of the multiplicand is multiplied by the multiplier. Hence, when a polynomial 
is divided by a monomial every term of the polynomial must be divided by 
the monomial. 

27. Divide 15 x 5 25 x* by -5 a 3 . Prove. 

28. (10 x 7 - - 8 a; 6 -f 3 x 4 ) -f- or 3 . Prove by multiplication. 

29. (a 2 ab ac) -. a. Prove by multiplication. 
27 X G - 3G x 5 cf + ab 

30. - - ol. 

9x a 

10 a 2 z -15 z 2 - 25s 8 



O/i. - * - OO. 

xy 

3 ab + 12 abx - 9 a 2 b a 8 -V6 

34. - ^- 35. 

Sab a 2 

36. ( 3 a 2 + f ab 6 ac) -r- -fa. Prove. 

37. (4 #y 8 ary + 6 ct'i/ 3 ) -i- - 2 a? t y. Prove. 

38. Make a rule for dividing a polynomial by a monomial. 

(1) How many terms of the dividend must be divided by the 

divisor ? 

(2) What rule have we for dividing one term of the dividend by 

the divisor ? ( 38, 14.) 

(3) How can the answer be proved correct ? 

39. (5 x?y s 40 a?a?y* + 25 a*xy) -. 5 xy. Prove. 

40. (4a6c-24o 2 & 2 -32aM+12a6)-j--4a6. Prove. 

41. ( S4aj 8 2/ 2 ~51a;y + 68iBy IT^)-?- ITsB 2 ?/. Prove. 

42. [3 a(x + ?/) + c\x + ?/) 2 - 5 b(x + ?/) 3 ] -=- (x -f y). See Ex. 19. 
A comment like that made in 35, 49, can be made here. 



54 THE ALGEBRAIC NOTATION 

43. To divide the product of several factors by a given quantity, 
divide one of the factors by the given divisor, leaving the other factors 
unchanged. 

Thus, (30 a 2 x 12 abc) + 6 a=5 a x 12 abc, or 30 a 2 x 2 be, = 60 a 2 bc. 

44. To divide the sum of several terms by a given quantity, divide 
each term by the given divisor, and add the quotients. 

Thus, (30 a 2 + 12 abc) -i-6a = 5a-f25c. 

Check the answers found in the four following problems by 
letting x = 3, y = 2 : 

45. (16 a,- 2 x- 24 ay 2 ) -5- 8 a;. 46. (16 a? - 24 a;/) -5- 8 a?. 

47. (3x 2 x$xy x6xif) + 3x. 48. (3 ar 5 + 9 it' y + 6 xf) -=- 3 a;. 

49. Divide a,* 2 -!- 11 a; + 30 by x + 6, comparing the method of 
solution with the arithmetical division of 675 by 25. 



SOLUTION, z 2 -f- 1 1 x + 30 

x 2 + Qx 



x + 6 divisor , 25)675(27 



x + 5 quotient 50 



5z + 30 175 

5 x + 30 175 

EXPLANATION. Let the student note how 675 is divided by 25 at the right. 
He will see that the first part of the dividend is divided at the start by the 
divisor 25. Then the divisor is multiplied by the quotient figure 2, and the 
product, 50, is subtracted from the dividend. Then, after another figure of 
the dividend is brought down, the remainder is divided by the divisor. 

In algebra, for convenience in multiplying, the divisor is commonly written 
at the right of the dividend and the quotient underneath it. The first term 
of the dividend, x 2 , divided by the first term of the divisor, x, gives the quo- 
tient, x. Hence, x is written as the first term of the quotient. Then the whole 
divisor is multiplied by the first term of the quotient, the product being placed 
under the dividend. Next x 2 + 6 x is subtracted from the dividend. The 
first term of the remainder is then divided by the first term of the divisor and 
the quotient 5 is written as the second term of the quotient. Last of all, the 
divisor is multiplied by the second term of the quotient and the product is 
subtracted from the remainder. PROOF. 

If the student will compare this multiplication care- x+6 
fully with the process of division, he will see that each x + 5 
line in the multiplication operation is likewise a line in x- -f 6x 
the division operation. Evidently the process of division + 5x4- 30 

is the reverse of multiplication. x' z + 11 x -f- 30 



DIVISION 



55 



Perform the division required in the following and prove by 
multiplication : 

51. or>+ 15 x + 



50. (.r- 

52. (r-llm + 30)-s-(m-5). 53. (3a? 2 + 10a;+ 3)-i-(a; + 3). 

- A / ' f^ ^ I ^^ I f~* *^\ / A.* I ^) n t\ f f 

C /\ / / /^t I _ / /^ II I * , ^ /** l * ( O* I " 1 1 \ ^\ T 

56. (4x 2 + 23:c + 15)H-(4a; + 3). 57. (6x- 7 x 3) + (2x 
58. 
59. 

60. (x* y*}-r-(x 
x-y 



61. (ac ad + Z>c 6d) -3- (a + 6). 



x-y 



x-y xy* 



xif-if 



ac 


ad -} 


- bc- 


bd 


a + b 


ac 


-4 


- be 




c d 




a<i 





bd 




ad 





bd 



b. As in arithmetic, terms of the dividend are not brought down until they 
are needed. Thus, in the first example, the term - - y 3 is not brought down 
from the dividend until it is needed on next to the last line. 



PROOF. o; 2 + xy + y 2 



PROOF, a + b 
c-d 



be 



x*y xif - 



ad bd 



62. 

64. x s _ 

66. 



ac -\-bc ad -- bd 
63. (2ac 6ad+5&c--15&cT)-*-(c 3d). 

65. (12 4 a 3 a 2 -f < 



. 67. (a 3 - 2 a& 2 + & 3 ) -s- (a - 6). 



56 THE ALGEBRAIC NOTATION 

68. Divide 20^- 5 a 4 + 50 ar> + 4^ + 2^-12 by 7x-2a? + x i 
-j- 5, and prove by multiplication. 

SOLUTION. In this problem the dividend and divisor should first be "ar- 
ranged." See 37. 



4 x 5 - 5 x 4 + 20 x 3 -f 50 x 2 + 2 x - 12 

4 x 5 - 8 x 4 + 28 x 3 + 20 x 2 



x 3 -2x 2 + 7x + 5 



4 x 2 + 3 x - 2 



2x 
15 x 



9x 2 --13x--12 
4x 2 - Use --10 



5 aj2 + x - 2 

To prove the work correct, we multiply the divisor by the quotient and 
add the remainder to the product. If no mistake has been made in either the 
division or proof, the sum will be the same as the dividend. 

PROOF, x 3 - 2 x 2 + 7 x +5 

- 8 x 4 -f 28 x 3 + 20 x 2 

- 2 x 3 + 4 x 2 - 14 x - 10 



4x 5 - 5 x 4 + 20 x 3 + 45 x' 2 + x -- 10 

5 x 2 + x - 2 

4x 5 -5x* + 20x 3 + 50x 2 + 2 x - 12 (Check.) 

Divide and prove : 

69. (2 a 4 + 3 a 2 - 5 a + 2 - 6 a 3 ) -r- (a 2 - 3 a + 1) . 

70. (6 s + 4 aW - 13 a 2 6 + 2 fr 3 ) -*- (2 a - 3 6). 

71. (2 m 4 + 6 m 2 - 4 ra - 5 m 3 -f 12) H- (m 2 - m + 1). 
72. 

73. 

74. (6 aV - 4 a 3 3 - 4 ax 3 + a 4 -f a 4 ) ~ (cr + x*-2 ax). 

75. x 6 - 



MISCELLANEOUS EXERCISE 57 

*76. (ax 2 - ab 2 + b~x - or) -=-(> -f b)(a - a;). 

77. (X G - a 6 ) -=- (ar* + 2 ax- 2 + 2 a 2 + a 3 ). 

78. (A^-faty + ttay + i^-Kfa + t!* 

79. Qr -hpg + 2|>r 2 g 2 + 7 gr ~ 3 r 2 ) -f- (p ~ g + 3 r). 

80. 

81. (a 4 - 16 & 4 ) -s- (a -2 6). 82. 

83. A 8 -A; 8 -r-/r + A- 2 . 84. a^ 1 - 



85. Divide 1 + 2 a; by 1 - - 3 a?, getting 5 terms in the quotient. 

Ans. 1 + 5 a; + 15 or + 45 a 3 + 135 a? 4 + etc. 

86. Divide 1 + 3 x 2 by 1 - - 4 #, getting 5 terms in the quotient. 

39. Miscellaneous Exercise in Addition, Subtraction, Multiplica- 

tion, and Division. 

1. From or 5 + 3 x 2 + 3 x + 1 take or* -- 3 a 2 + 3 a; - - 1. Check 
with x = 2. 

2. Add3 ajy - 10 y\ - x 2 / + 5 y\ 8 afy 8 - 6 /, 3 a? 4 + 4 afy 2 + 2 ?/ 4 . 

3. Divide 8 ar 3 + / by 2 a? -f 2/- Prove. 

4. (a 2 4- aa? + a; 2 ) (a 2 - ax + or) = ? 

5. (a^ + 6c)a 3 6 = ? 6. 4 a 3 6a? x 7 ftV. 7. 
8. (a 4 6 s a; - 5 6 fl ) X i oV. 9. (a; 2 + 2 a 

10. 12^ 5 -9^ 3 + 6.T 2 --3x 2 . 11. 
12. 

13. (ab + cd + ac + bd)(ab 

14. From -I a 2 - 4 a 1 take a 2 + a . Check with a = 4. 

O ^5 O * 

15. What expression must be added to 5 x 2 - 7 x + 2 to produce 



58 THE ALGEBRAIC NOTATION 

16. Subtract 3 3? - 5 x + 2 from 1 and add 5 x 2 6 xto the result. 

17. Multiply |- #?/ by - 3 # 2 -f |- or?/. Check with sp = 6, y = 1. 

18. (8 x 8 x 2 4- 5 x 3 -f- 7) -r- (5 a? 3). Prove work correct. 

19. (30x + 9-71x* + 2Sx 4 -3ox 2 ) + (x*-13x + 6). Prove. 
* 20. Divide (3 x 4 - I + 3 x + 6 x 2 + 7 or 3 ) (1 + a; 2 - a) by a; + 1 + or. 

21. Divide 8 on/ 2 + 3 # 4 5 a 1 ?/ 3 7 # 3 ?/ + ?/ 4 by x 2 -j- 2/ 2 #?/, 
arranging first according to descending powers of x\ then divide, 
arranging according to the descending powers of y, and compare 
the quotients. How are the quotients alike? How different? 

22. 5 (a - &) + G(a - b) - 4(a - 6) = ? 

23. 3(3* - /) + 7(0* - r) - 2(ar - /) = ? 

24. Simplify 2a-2(6 + 3a) -3{6 + 2(a- &)}. 

In working this and the following problems the student must remember that 
the absence of sign between quantities denotes multiplication and that the order 
of performing operations, as stated in 27, is powers and roots first, multipli- 
cation and division next, and addition and subtraction last of all. However, 
quantities inclosed in parentheses must be treated as single quantities. 

SOLUTION. 

2 a - (2 b + 6 a) - 3 (6 + (2 a - 2 5)}. (Multiplication performed first.) 

2 a 2 b 6 a 3 {2 a 6}. (Parentheses removed and terms collected.) 

2 a - 2 b - 6 a - (6 a - 3 6}. (Multiplication first. ) 

2a 26 G a 6a + %b = b 10 a. ^.?is. (Simplification.) 

Simplify the following: 

25. a [56--{a--3(c 6)4-2c (a26 c)J]. '-4ns. a. 

First perform the multiplication inside the brace ; then remove the brace ; 
next remove the bracket ; lastly simplify. 

26. (x + y)(y + z) (z + u)(u + x) ({B + z)(y u). Ans. y z --u 2 . 

27. \x(x-\-a) a(x-}-a)l {x(x a) a(a x)\. 

Ans. a 4 - 2 a 2 .T 2 + a 4 . 

28. (a? + 6) (a? + c) - (a + b + c)(x + 6) + a 2 + aft + b 2 + 3 



2 a# + a 2 . 
29. a + & 2 -a + &a-Z>-2&-2-& 2 -2a. -4wa. 36 2 . 



MISCELLANEOUS EXERCISE 59 

40. Miscellaneous Exercises in Fundamental Operations to be 

solved by inspection, that is, mentally. 

1. From 16 bx 2 take - - 3 bx 2 . 2. Add 6 m z p and 5 m 2 p. 

3. From 3x 2y + 5z take 5 y z 8 x. 

4 Q r 3 fi - 8 ri*h fi fi r/ 5 r 6 ?/ 9 7 

Tt. t/ t// t/ C/ vt \J O. \j \JL *v u I . 



Subtract: 

8. 4 a; 9. -6 a 2 10. 6c 11. 1 y?y 

11 x 4 a 2 56c -12ar.y 

12. -48 -T- -4. 13. 21 a 10 -f- 3 a 7 . 14. -70a 7 6 8 --14 

Add : 

15. 5- 76 16. - 6^--16?/ 2 17. -17c/w 2 + 7 

4 a + 12 6 9 ar 5 + 3 y 

-2 a- 66 



18. If a (b c) = ab ac, what does 34 (50 - - 1) = ? What is a 
good way to multiply 34 by 49 ? Multiply in same way 26 by 39 j 
45 by 78 j 120 by 151. 

19. (5z + 12)7z. 20. 3m(10m 2 -4?0- 21. 2 a (6 a 2 -3a-f-4). 
22. (25 a 8 - 15 a 6 ) --5 a 2 . 23. (54 4 & S -GO 7 //'+6a 4 6 4 )--6a 4 6 4 . 
24. (5a;-7)(3a?H-2). 25. (2rt-3)(6a-7). 

26. 9 w +(- - 4 m + 6 ?i) - (3 m - - n). 
SUGGESTION. Add m's first ; then w's. 

27. 2a?-3- 
28. 

29 . (5 m + 9 n + 4 a;) + ( - 3 x 7 // - 6 n ) - (10 y - - 8 a? - 2 m) . 

30. Subtract 12 # - - 7 ?i 6 // from 11 n + 3 m 8 a?. 

31. 5 x 9 / x 9 ?/V x ass 7 . 



60 



THE ALGEBRAIC NOTATION 



* 32. - 3 a% x - 5 b s c x - 2 c 3 X - aW. 

33. (6 a-V - 9 ab 2 x 2 - 15 a W) -f- - 3 ax*. 

34. m# 2 - - [8 y -\- (6 a mx) 2 a]. 



35. 5f a 2 m 2 -(-2i 

36. Sma-S^ + S 

37. 3 a 2 - 6 ab - 8 6 2 -f- 7 a 2 - 3 a 2 + 2 aft - 14 W - 6 06. 

38. 4 a 2 - 6 a + 4 - 3 a 2 + a + 1.5 a 2 - 2 - 3.4 a 2 - 3.75. 

39. Prom #*,.+ 1 take 1 2 x + # 4 + 3 x 2 4 or 5 . 

40. Add 6 n, In, 3 n } 18 w, and 11 n. 

41. 4a-2&- 



41. Review of Definitions. The best definitions for the student 
are those which he makes from his own knowledge of words. 

Let the student write out in his own language from the knowl- 
edge he has gained of the thing defined, and not by looking up 
the definitions given in the book, the definitions of the following, 
and bring them to class for discussion : (a) Addition in algebra. 
(6) Subtraction in algebra. (c) Multiplication. ( d) Division. 
(e) Exponent of a quantity. (/) Coefficient, (g) Literal part. 
(7i) Term, (i) Similar terms, (f) Quantity. (k) Polynomial. 
(I) Power of a quantity, (m) Eoot of a quantity, (n) Positive 
quantity, (o) Leading letter, (p) Distributive law. ((/) Symbols 
of aggregation. (?) Equation (see equations, p. 37), etc. 



61b. 




6-2 Ib. 



6-2 Ib. 




An equation ( 31, 42) resembles weights balanced on scales, 
In the figure we see equal weights from equals leave equals. 



INTEGRAL EQUATIONS Gl 

INTEGRAL EQUATIONS 

42. Solution of Integral Equations. An equation is a statement 
that two quantities have the same numerical value. 

Each problem in 1 and 31 led to an equation. In the solu- 
tion of the equations of those articles, the answers were reasoned 
out. We will now proceed differently. From the example below 
we can see that any quantity may be "transposed" (carried across) 
from one side of an equality sign to the other by changing its sign. 

Thus, 6 x + 44 = 4 x + 56 (By subtracting 44 + 4 x from each 

becomes 6x 4 cc = 56 44 side of the first equation. Equals 

or, 2x = 12 subtracted from equals leave equals.) 

whence, x = 6 (If 2 x = 12, x = \ of 12.) 
PROOF 

When x = 6, 6a; + 44 = 4x + 56 

becomes 6x6 + 44-4x6 + 56 
or, 80 = 80 

To solve an equation, "transpose" so that all terms containing $ 
are on the left side of the equality sign, and all terms which do not 
contain x are on the right side. TJien simplify each side. Last of 
all, find x by dividing the right " member " of the equation by the 
coefficient of x. (If coefficient is negative, say 4, divide by 4.) 

Prove the answer right by substituting it in the given equation. 

1. 11 + 10 = 8 +-19. 2. 7x + l = 4 a; + 13. 

3. 16a?-ll = 7a? + 70. 4. 24^-49 = 19 x- 14. 

5. 23 x- 50 = 18 x -15. 6. 3 a + 24 = 79 -2x. 

7. 16a; = 64-(12-3aj). 8. 8x-7 -3z = 2-4 x. 

9. 5 (x 7) + 63 = 9 x. SUGGESTION. First multiply x 1 by 5. 

10. 16aj = 38-3(4-o;). 11. 9(7- a) =6- 3(5 -a). 

12. 4(aj-2) = 2a;-7(aj-4). 13. 6o?H-4#- 16 -2a; = 0. 

14. Find a number which added to 22 less than itself = 216. 



62 



THE ALGEBRAIC NOTATION 



15. A man who had paid out $700, $400, $200 respectively 
towards the education of his three sons, directed in his will that 
this difference should be equalized in the settlement. If his 
property was worth $ 9200, how much did each son get ? 

SUGGESTION. Let x dollars, (x + 300) dollars, (x + 500) dollars equal 
sums the sous got. Why ? 

16. A stock fence has nine 

4 + 5 a? wires, the lowest three spaces being 
3, 3, 4 in. respectively, and the 
others growing wider upward by 
the same amount. Find the dis- 
tances apart to put the wires if the 
fence is to be 52^ in. in height. 



4 + 305 



17. How many cents a pound is paid for butter when 
added to the cost of 2 Ib. gives the same sum as 2/ added to 
the cost of 4 Ib ? 

18. A cashier paid out $26 in $1 and $2 bank notes, and the 
number of notes of the latter kind was less than the number of 
the former kind by 5. How many notes of each kind did he use ? 

19. A boy has 155 hills of potatoes to dig in 5 hours. He pro- 
poses to lay out his work in such a way that during each hour he 
may have 6 hills less to dig than during the hour preceding. To 
do this how many hills must he dig each hour ? 

3t x = number dug first hour. Then how many were dug 
The third ? etc. 

20. Using round numbers, there were in four of the largest 
libraries in the world, in 1909, 8| millions of volumes (including 
pamphlets). The Koyal Library in Berlin had ^ million less 
than the Library of Congress at Washington and 1 million less 
than the British Museum in London, and the Bibliotheque Rationale 
in Paris had as many as both the Berlin Library and the Museum. 
How many volumes had each ? 

21. In 1900 had the white population of continental United 
States been reduced by 4,621,238, there would have been 7 white 



SUGGESTION. 
the second hour ? 



INTEGRAL EQUATIONS 63 

persons to every one of negro blood. The population, exclusive 
of Chinese, Japanese, and Indians, was 75,693,190. What was 
the white population, and what was the negro population ? 

22. In the 573 colleges and universities of the United States in 
1906-1907 there were 161,760 students. Of this number the male 
students were 2,685 less than 2 times the number of female 
students. How many were there of each? 

23. In the colleges and universities of the United States in 
1907-1908 there were 24,489 professors and instructors. Of these 
there were 1634 more than 4 times as many male teachers as 
female teachers. How many were there of each? 

24. According to the census of 1900 the total population of 
Ohio, Illinois, and Pennsylvania was 15,281,210. The population 
of Illinois was 664,005 greater than that of Ohio, and the popula- 
tion of Pennsylvania was 1,480,565 greater than that of Illinois. 
What was the population of each ? 

25. In 1906-1907 there were 367,036 more than 21 times as many 
pupils in the public elementary schools of the United States 
as there were in the public high schools. In both there were 
16,890,818. How many were there in the secondary schools, and 
how many in the elementary schools ? 

26. The sum of the three dates (all A.D.) on which Jerusalem 
was destroyed by Titus, the Roman Emperor Constantine recog- 
nized Christianity, and Constantinople was taken by the Turks, is 
1836. From the first date to the second is 243 years, and from 
the second date to the third, 1140 years. What is each date? 
Prove answers. 

27. The heights of the Eiffel iron tower in Paris, the Washing- 
ton monument in Washington, D.C. (each the highest in its class), 
and the twin spires of the Cologne cathedral, added together, give 
a total of 2051 ft. The Washington monument is 43 ft. higher 
than the cathedral, and the Eiffel tower is 40 ft. less than twice 
the height of the cathedral. What is the height of each ? 



64 THE ALGEBRAIC NOTATION 

28. New York City has the highest office buildings in the 
world. The total height of three of them, the Flatiron, World, 
and Times buildings, is 1081 ft. The World building is 89 ft. 
taller than the Flatiron, and the Times 45 ft. taller than the 
World building. What is the height of each ? 

29. The sum of the heights of the Metropolitan Life and 
Singer buildings is 1312 ft. and their difference is 88 ft. What 
is the height of each ? 

30. The wheat crop of the world in 1906 was approximately 
3423 millions of bushels. The number of bushels produced in 
other countries was 483 millions of bushels more than 3 times 
that produced in the United States. How many bushels were 
produced in the United States? 

31. In 1908, among our 464 colleges, universities, and technical 
schools for men and for both sexes, the N. Atlantic states had 
twice as many as the Western, 55 less than the Southern, and 
101 less than N. Central states. How many had each region ? 

32. Of the 464 institutions of Ex. 31, Ohio had the largest num- 
ber, having 7 more than Pennsylvania and 10 more than New 
York. Four times the number in these three states and 9 more 
equals the number in all the other states. What was Ohio's number ? 

43. Historical Note. Algebra is not so old a science as either 
arithmetic or geometry, though it goes back farther than the 
Arabic notation of arithmetic. The earliest recognized writer on 
algebra was Diophantus of Alexandria, Egypt, who died about 
330 A.D. It should be said, however, that Annies, an Egyptian 
writer, who lived about 1700 B.C., wrote a book that contained 
the solution of some simple equations. His work, found on a 
papyrus now in the British Museum, was translated by Eisenlohr 
in 1877. Diophantus wrote his work, called Arithmetica, in the 
Greek language, and introduced the idea of an algebraic equation 
expressed in symbols. Apparently he did not clearly understand 
the nature of negative numbers, which he avoided. 



HISTORICAL NOTE 65 

After Diophantus there came the following eminent writers on 
the subject : Aryabhatta (about 500 A.D.), Brahmagupta (about 
630 A.D.), and Bhaskara (born 1114 A.D.), all Hindu writers living 
in India. Following Aryabhatta and Brahmagupta came Arab 
authors, prominent among whom was Mohammed ibn Musa, gen- 
erally called Al-Chwarizmi. Europe got its knowledge of algebra 
from Arabia, and Europe communicated the knowledge to the 
rest of the world. More will be said later of the persons just 
named. 

It is interesting to find that there have been three stages in the 
development of algebra: (1) the rhetorical stage, in which all 
solutions were written out in full, much as in arithmetical 
analysis ; (2) the syncopated stage, in which abbreviations for 
important words were introduced; and (3) the symbolic stage, 
in which symbols are used to denote numbers, operations, and 
relations. Ahmes and the earliest Italian and East Indian 
writers used the rhetorical form. Thus Metrodorus, a Greek 
writer of about 310 A.D., Aryabhatta, and Brahmagupta used the 
rhetorical form. Diophantus, Al-Chwarizmi, and Bhaskara, on the 
other hand, used the syncopated form. This syncopated algebra 
was used up to the sixteenth century. From then on algebra be- 
came symbolic in form. During the sixteenth and early part of 
the seventeenth centuries writers were introducing one symbol 
after another. Thus Widmann in Germany introduced + an( l 
- ; Robert Recorde in England introduced =, and Oughtred, X ; 
and so on. The present symbolic algebra in which symbols are 
almost exclusively employed is considered to have been introduced 
and used first by Vieta (1540-1603) in France. Since Vieta's day, 
algebra has been developed by a long line of students and in- 
vestigators, prominent among whom may be named Sir Isaac 
Newton in England, Descartes in France, and Euler in Germany. 



CHAPTER II 



AB 



THEOREMS IN MULTIPLICATION AND DIVISION. FACTOR- 
ING. LOWEST COMMON MULTIPLE 

44. A theorem in algebra is a rule or law expressed in words. 
A formula is a rule or law expressed in symbols, 

45. Theorems in Multiplication. 

1. Let A and B represent any two algebraic terms, or two lines. 
Then A +B 

A B A +B 

A A* AB A 2 + AB 

+ AB + B 2 
A 2 + 2 AB + B 2 

From. the multiplication above and also 
from the diagram, we see that : 

(A + B)' 2 = A 2 + 2 AB + B 2 . 

(A + By 2 means " the square of the sum of any two quantities, 
A and B, of which A is the first and B the second. 1 ' 

A 2 + 2 AB + B 2 is " the square of the first quantity, A, + twice 
the product of the two quantities, A and B, + the square of the 
second quantity, B" Or, we may write 

(1st number + 2d number) 2 = (1st number) 2 + 2(lst number) x 

(2d number) + (2d number) 2 . 

Changing the formula (A + B) 2 = A 2 + 2 AB + B 2 into a theo- 
rem, we have 

THEOREM I. The square of the sum of two quantities is equal to 
the square of the first, plus twice the product of the first quantity mul- 
tiplied by the second, plus the square of the second. 



THEOREMS IN MULTIPLICATION 67 

a. REMARK. Notice how much shorter the formula is than the theorem. 
In the formula there are only 16 marks, counting each letter and sign as a 
mark. In the theorem, counting each letter and comma as a mark, there are 
144 marks, or nine times as many as in the formula. In this example we see 
how wonderfully compact the algebraic notation is. 

2. Square the sum of 7 and 4 by the theorem and check result, 



- 49 + 5G + 16 = 121 
or, II 2 =121. Check. 

3. (3a 2 + 2m) 2 = ? 

SOLUTION. (3 a 2 + 2 wi) 2 = (3 2 ) 2 + 2(3 a 2 x 2 m) + (2 n) 

= 9 a 4 + 12 a' 2 m + 4 m 2 . Ans. 
PROOF. 3 a 2 + 2 m 
3 a 2 + 2 m 



9 a 4 + ti a-??* 

+ 6 g?m + 4 m 2 
9 a 4 + 12 a-w + 4 m 2 

Write the results directly by use of the theorem. Check by 
actual multiplication until the use of the theorem is well un- 
derstood. 

4. (4cr-f9w) 2 . 5. (2 a 2 + 56)". 6. (5w+3n)(5m+3w). 

7. (a + 5)*. 8. (2a + 4) 2 . 9. (a + 7 6)(a -f-7 6). 

10. (7z + 3c) 2 . 11. (a 8 + 66)*. 12. (3 a 2 + 5 6 s ) 2 . 

13. (l+2a s 6) a . 14. (JoJ + ^y) 2 . 15. (6 + 10 a- 2 ) 2 . 

16. 21 2 = (20 -f I) 2 . 17. 72 2 = (70 + 2) 2 . 18. 102 2 = (100 + 2) 2 . 

19. 5i 2 = 5- 2 . 20. 12> 2 =12+i 2 . 21. 



^4 7? 

22. Let ^1 and 5 again represent any two 

terms. Then from the multiplication in the - 



margin, we have _ AB 4- 



68 THEOREMS IN MULTIPLICATION AND DIVISION 

Changing this formula into a theorem, we get 

THEOREM II. The square of the difference of two quantities 
is equal to the square of the first, minus twice the product of the first 
quantity multiplied by the second, rjlus the square of the second 
quantity. 



23. (^ 

SOLUTION, (a - 3 &) 2 = a 1 - 2 (a x 3 ?>) 4- (3 6) 2 

= a 2 - 6 ab + 9 b*. Ans. 
PROOF, a 3 b 

a-3b 



a? 3 ab 

- 3 ab + 9 62 
a 2 - a& + & 2 

Using Theorem II, write directly the answers to the following : 
check by actual multiplication until the application of the theo- 
rem is well understood. 



24. (.-4?/) 2 . 25. (2x-3y) 2 . 26. (x-2m)(x-2m). 

27. (p*-3q)' 2 . 28. (5z 2 -4) 2 . 29. (4 a - 3) (4 a - 3). 

30. (r-f) 2 . 31. (2.r 2 -5?/ 4 ) 2 . 32. (3 3 -6)(3 a 3 - 

33. (l-8afy) 2 . 34. (2-i&) 2 . 35. (Ja - J&)(a 

36. 99 2 =(100-1) 2 . 37. 147 2 = (150-3) 2 . 38. 995- = (1000 -o) 2 . 

39. Let A and J5 represent any two terms. A + B 
Then from the multiplication in the margin, we A B- 
see that A 2 + AB 



Changing this formula into a theorem, we get 

THEOREM III. The product of the sum and difference of two 
quantities is equal to the difference of their squares. 

REMARK. Notice that (A + B} (A-- B} is the product of the sum and 
difference of two quantities of which A is the first and B is the second. No 
sign between the parentheses indicates a product. 



THEOREMS IN MULTIPLICATION 69 



40. 

SOLUTION. (3 a + 5 6) (3 a - o 6) = (3 a)- - (5 b)- = 9 a 2 - 25 6-. Ans. 

PROOF. 
Write the answers to the following di- 3 + 5 6 

rrctly by use of Theorem III. Check by 3a - 

actual multiplication until the use of the ; ^ ^ _ 2 - fca 

theorem is well understood. 9 a 2 _ 25 b- 






41. (5a + 26)(5a-2&). 42. (3 a + 11 &)(3a- 11 

43. (3 m 2 - 4 p)(3 m 2 + 4. 44. (11 m 2 + 6 n 2 )(11 ??i 2 - 6 n 2 ). 

45. (4 3 - 3 6 3 )(4 a 3 + 3 6 3 ). 46. (3 .r/ - 7 .^/)(3 x/ + 7 

47. (im 8 + l)am s -l). 48. 

49. (a 5 -l)(a 5 + l). 50. 

51. (a& + cd)(a6-cc?). 52. 

53. (5a 8 6 2 + 2aj)(5a 8 6 2 -2a;). 54. 

55. 18x22=(20-2)(20 + 2). 56. 49 x 41 = (45 + 4)(45- 4). 

57. 103x97 = (100 + 3)(100-3). 58. 1025x975. 

59*. (24 + 4)(24-4) = 24 2 -4 2 = 28x 20=560; then 24 2 = 560 
+ 16 = 576. 

60*. Square in the same way 33 ; 72 ; 124 ; 117, efc. Check by 
actual multiplication. x 4- a 

61. Let .r, a, and 5 represent any three x + b 
terms. From the multiplication in the &~ + # 
margin, we see that + bx -\-ab _ 

(x + a)(x + 6) = or 2 + (a + 6)^ + a6. x> + (a + b)x + ab 

Changing this formula into a theorem, we have 

THEOKEM IV. The product of two binomials having a common 
term equals the square of the common term, and the algebraic sum of 
the other terms times the common term, and the algebraic product of 
the other terms. 

* See 1, p. 6. 



70 THEOPtEMS IN MULTIPLICATION AND DIVISION 

62. (a? + 6) (a? - 9) = ? 

SOLUTION, (x + 6)(x + [ - 9]) =x 2 H- (6 -f [ 9])x + 6 x - 9 

= x 2 3 x 54 PROOF. 

x -4- 6 
SUGGESTION. Whenever a second term is negative, 

write it in a bracket preceded by +, as in the problem 



ff2 I (j y- 

just solved, until the process is well understood; afterward 

this will not be necessary. = ~ 

y? 3 x 54 

63. (a? + 2)(a;-6). 64. (a; + 5)(a;-ll). 

65. (a; + 6) (a; + 9). 66. (a; + 7) (a; + 8). 

67. (m-2)(m + 10). 68. (a; -8)(a?- 12). 

69. (z 4 -3)(V-9). 70. ( + i)(-i). 

71. (a 2 -5)(a 2 -7). 72. (a 3 + 5)(a 3 - 7). 

73. (a 2 + 4~)(V + 5). 74. (aj 2 4)(a? 2 -5). 

75. (/-4)(2/ 2 + 5). 76. - 

77. (ojy-4)(ajy + 16). 78. 

79. (5o-36)(5a-76). 80. (3o-6)(3a 

81. (3-aj)(4-a?). - 82. (5 + 6 2 )(7 + 

83. (2z + lT)(2z-13). 84. (2m 3 -ll)(2m 3 -12). 

85. (3 x 2 - 7)(3 ^ 2 + 16). 86. (4m 4 - 15) (4m 4 + 5). 

87. Let A and B represent any two 
algebraic terms. Then, from the mul- 

A ~D 

tiplication in the margin, we have 



A* AB 
2 B\ 



Put into a theorem, this formula be- A 2 - 2 AB + B 2 
comes A - B 



THEOREM V. The cube of the differ- A * - 2 A'B + AB 2 
ence of two quantities equals the cube of A B + -, AB- - B 

the first, minus three times the square of A* 3 A-B + 3 AB 2 B 3 



THEOREMS IN MULTIPLICATION 71 

the first times the second, plus three times the first times the square 
of the second, minus the cube of the second. 

a. Evidently in the cube of the sum of two quantities, all the signs on the 
right side of the formula are positive. 



88 . ( a s _2 by = ? Expand " and check with a = 2, 6 = 1. 

SOLUTION, (a* - 2 &)' = (a 2 ) 3 - 3(a 2 ) 2 (2 6) + 3(a 2 )(2 ft) 2 - (2 6)3 

= a e ._ (3 4 & + 12 a -b* - 8 6^ ( 27) 

(4 _ 2)3 = 2* r= 64 - 96 + 48 -- 8 = 8. Check. 



89. (a 2 -36) 3 . 90. (?ft-4w) 8 91. (a?- 

92. (a 2 + 2/) 3 . 93. (a 2 + 66 3 ) 3 . 94. (2a s -6) s . 

46. Miscellaneous Exercise in Applying the Theorems of Multi- 
plication. 

As a valuable exercise let the student point to the figures, let- 
ters, and signs in the formulas, while he says the corresponding 
words of the several Theorems I-V. 

1. (ra + ?i) 2 . 2. (o + 7)(a-9). 

3. ( + ra)(a--w). 4. (2a--6) 2 . 

5. (2a-36)(2a + 36). 6. (c-7 y)(c+3y). 

1. (a 2 -ft 2 ) 2 . 8. (ax + by)(ax-by). 

9. (3 xy - 8)(3 xy - 7). 10. (16-aj) 2 . 

11. (m 3 -3n 2 ) 2 . 12. (2 a*x + 3 % 3 ) 2 . 

13. (c-/0*. 14- 

15. (4a + 7x 2 ) 2 . 16. 

17. (6jj -5 ^(6^ + 5g). 18. (*.B- 

19. (4m 2 -3?i 8 ) 2 . 20. (a 2 -7) (a 2 + 12). 

21. (a 8 -7c)(a s + llc). 22. (3 a 2 -I) 2 . 

23. (3^ + 2/ 2 )(-^/ + r)- 24. (,.y-a)(^r + 

25. (^r 2 -is 3 ) 2 . 26. (2 a 4 - 11) (2 4 + 15). 

27. 3a^-5 2 3x 2 -f 7?/ 2 . 28. 



72 THEOREMS IN MULTIPLICATION AND DIVISION 

29. Find the square of 7J by Theorem I, writing (7) 2 =(7+i) 2 . 
Check by squaring - 2 ^. 

30. (5l) 2 ; (41) 2 ; (201) 2 ; 41 2 = (40 + I) 2 ; 52 2 ; 79 2 =(80-1) 2 ; 
99 2 . Cheek the answers by ordinary multiplication. 

31. Find the product of 42 and 38, that is of 40 + 2 and 40-2. 

SOLUTION. (40 + 2) (40 - 2) = 1600 -- 4 = 1596. Check by actual mul- 
tiplication. 

32. 48 x 52. 33. 81 x 79. 34. 93 x 87. 35. 95 x 105. 

36. 88JX91J. 37. 257x263. 38. 546x554. 39. 125x135. 

40. (ar 5 -l)(ar 5 +l). 41. (a*-7b) 2 . 42. (a-b) 3 . 43. (5|-) 3 . 

44. 999 2 . 45. (6.5) 2 . 46. (101) 2 . 47. 242x238. 

47. Theorems of Division. 

1. Since (A + B)(A-B) = A 2 -B* by 45, Th. Ill, it fol- 
lows that 



and 

Stated as a theorem, these formulas give 

THEOREM VI. The difference of the squares of two quantities is 
exactly divisible by either their sum or their difference. 

Or, The difference of the squares of two quantities equals the 
product of the sum and difference of the quantities. 



2. (A 3 -B S ) + A-B)=? 3. 

SOLUTION SOLUTION 



A* - 
A 9 - 



A-B 



A* + AB + B* A* + A-B 



A + B 



A* -AB + B 2 



A 2 B - A-B 

A 2 B - AB 2 - A*B - AB 2 



AB 2 -- B* + AB- + 

AB 2 - B s + AB 2 + 



Thus, (^l 3 -&)-*- (A - B)=A 2 -\-AB+B 2 ; 



THEOREMS IN DIVISION 73 

Stated as theorems, these formulas give 

THEOREM VII. The difference of the cubes of two quantities di- 
vided by the difference of the quantities equals the square of the first, 
pi as the product of the first by the second, plus the square of the second. 

Or, The difference of two cubes equals the difference of the quan- 
tities times the sum of the square of the first, the product of the first 
and second, and the square of the second. 

THEOREM VIII. The sum of the cubes of two quantities divi<l<-'! 
by the sum of the quantities equals the square of the first, minus the 
product of the first by the second, plus the square of the second. 

Or, The sum of two cubes equals the sum of the quantities times 
the square of the first less the product of the first and second, plus 
the square of the second. 

a. REMARK. The student should notice particularly the sign of the second 
term of the quotient; that is, when it is + and when . He should con- 
trast a 2 + 2 ab + b' 2 with a 2 + ab + b 2 . The first is the square of a + b 
( 45, Th. I) while the second, as we have just seen, is the quotient of a 3 b B 
divided by a b. He should also contrast (a + 6) 3 and a 3 + & 3 , and (a &) 3 
and a 3 6 3 . See 45, Th. V. Confusing these is a common mistake. 

In the following exercises tell what each quantity is divisible 
by and what the quotient is when so divided. Check results by 
actual multiplication. 

4. a 3 y 3 . 

SOLUTION, x 3 ?/ 3 is divisible by x y. The quotient is x 2 + xy + y 2 . 

PROOF, x 1 + xy + y 2 

x -y 

x s + x 2 y + xy' 2 

x 2 y xy 2 y 3 
x s y 9 

5. m 3 ?i 3 . 6. ??i^ + 7i 3 . 7. x' y-z". 8. .T^ + T/'V. 

9. x>-y\ 10. a 3 + (2c) 3 . 11. x^-(3y)\ 12. (2m) 2 - (on) 2 . 

13. 8 or 5 - 27 f. 

SUGGESTION. 8 x 3 - (2 x) 3 , and 27 y* = (3 ?/) 3 . Hence 8 r, 3 27 y* is divis- 
ible by 2 x - 3 y. The quotient is (2 x) 2 + (2 x) (3 y) + (3 y) 2 , or 4 x 2 + 6 xy + 
9 y 2 . Prove by multiplication. 



74 FACTORING 

14. 8^-125. 15. 8 3 + 125c 3 . 16. 

17. a 3 + 125. 18. -216 -if. 19. 9x 2 y 2 -25z 2 . 

20. 64 + /. 21. 64 a 3 -343. 22. 64 or 3 -!. 

23. 64 a 6 + 125 ?/ 9 . SUGGESTION. The divisor is 4 2 +5 y s. 

24. z 3 _64?/ 6 . 25. x + 27y 3 . 26. 27m 9 -64^. 
27. 216j9 12 -9 6 . 28. 8z 12 + l. 29. 343 z 3 - 1000 y> 
30. A 2 + B 2 + A + B. 31. 



Show by long division that divisors in Ex. 30, 31 are not exactly 
contained in the dividends. Make theorems therefrom. 



32. (A 3 -B*) + (A + B). 33. (^l 3 + B s ) -=- (A - B). 

Divide and make theorems as in the preceding exercises. 

FACTORING 

48. Factoring in algebra is separating a quantity into others 
which multiplied together will produce the given quantity. The 
same definitions are used in algebra as were used in arithmetic. 
Thus , SL prime quantity is exactly divisible only by itself and one. 

We will take up in the following articles only a part of the 
various cases in factoring; others will be given later on. 

49. Factoring of Monomials. A monomial, such as 5 3 6 2 c, is re- 
garded as being already factored. A monomial, such as 72 xy 1 , is 
factored when its coefficient 72 is factored. Thus 72 xy' 2 =2 3 3 2 xy 2 . 

50. Monomial Factors in Polynomials. A monomial is a factor 
of a polynomial when it is contained in every term of the polynomial. 

Type forms : 

ax + bx == (a + b)x ; 3 mx + 3my 3mz = 3 m(x +y z). 



MONOMIAL FACTORS 75 

1. Factor 8 a 3 2 ax and prove answer by multiplication. 

DIVISION SOLUTION. 2 g)8 a 9 2 ax PROOF. 4 - - x 

4 a 2 x 2 a 



DIRECT SOLUTION. 8 a 3 2 ax = 2 a (4 a 2 #) 8 a ! - 2 aaj 

Factor the following and prove the answers correct by multi- 
plying the factors together: 

2. 4 a 2 2 a. 3. 25 m 2 w - - 5 w. 4. or 3 x 2 y. 

5. a--atf. 6. 3 or + or 5 . 7. 15a 2 -75ee 4 6 2 . 

8. 18 -81 a. 9. 3 a 3 -3 a. 10. 9x-y 2 -9xy 2 . 

11. 3 a 4 - 3 a 3 6 + 6 a 2 b 2 . 
DIVISION SOLUTION. 

3 a 2 )3q-3q 8 6 + a-6 2 PROOF, a 2 - ab + 2 6 2 

a 2 - a& + 2 Z> 2 3 a 2 



3 a 4 - 3 a 3 6 + 6 a 2 6 2 
DIRECT SOLUTION. 3 a* - 3 a*b + 6 a 2 6 2 = 3 a' 2 ( 2 - a^> + 2 & 2 ) . 

12. 35 a - 21 a 3 + 14 a 4 . 13. 3 m 5 - 12 m s n 2 + 6 mn 4 . 

14. 2ar 3 + 6ari/ + 2a 2 . 15. 4 ./ 3 - 4 ary 2 + x 3 ?/. 

16. 8z 4 -4x 2 -4. 17. a 3 - 6 orb 4- 10 aft 2 . 

18. Make a rule for factoring a polynomial that has a mono- 
mial factor in it. How is the monomial factor found ? Should it 
contain every factor which is contained in each of the terms of the 
given polynomial ? How is the polynomial factor found ? How 
is the answer proved correct ? 

19. 63Z 3 54#. 20. wV + w 3 w 4 --w?i 6 . 
21. 12 ax 2 - 3 bx~ + x 2 . 22. tfy* - x*y* + x*y\ 

23. 14 be 4 - 21 b 2 c s + 7 b 3 c 2 x. 24. a 6 - 5 a 5 - 2 4 + 3a d . 

25. 3a 3 -9a 2 6 + 9a& 2 -36 3 . 26. 24 Z> 4 - 36 6 3 + 185 

27. x 12 + x 11 x 10 x 9 . 28. ac be cy ~ abc. 
29. 



SOLUTION. 6 _+ c| q(6 + c)- 2 m(5 + c) 

a 2w 



76 FACTORING 

30. 5a(2x-y) + 3b(2x-y). 31. 3 x*(x - 1) - 5(x - 1). 

a. Throughout factoring the first step to be taken is to remove monomial 
factors from given expressions if any are present. If the student does not see 
how to begin the solution of an exercise, the reason will very often be that the 
problem is disguised by the presence of a monomial factor. 

BINOMIALS 

51. The Factoring of Binomials. Four different cases or classes 
of problems are taken up in this chapter. 

I. Binomials that are the Difference of Two Squares. 
Type form: a 2 -b 2 = (a + b)(a -b). 

By 47, Th. VI, the difference of the squares of two quantities 
equals the sum of the quantities multiplied by the difference of 

the quantities. 

PROOF 

1. Factor 4 a 2 b 2 and prove by actual multi- 2 a + b 

plication. 2a b 

4 a 2 + 2 ab 
SOLUTION. 4 a 2 - 6 2 = (2 a) 2 - 6 2 = (2 a + 6) (2 a - 6) . - 2 ab - b 2 

4 a' 2 ~^b* 

Factor and prove the following : 

2. 9a 2 -6 2 . 3. or 9 -4. 4. x 2 -16y 2 . 5. 25 -c 2 . 
6. a 2 x 2 y 2 . 7. m 2 n 2 . 8. a 2 9. 9. m 2 1. 

10. 9a 4 -166 2 . Ans. (3a 2 + 4ft)(3a 2 -4 6). 



Factor the following, proving the first three or four (and 
any others the student does not feel sure are correct), by actual 
multiplication : 

11. 4m 2 -25n 2 . 12. 16m 2 -49. 13. 9# 4 -49ij/ 4 . 

14. a 2 x~-35y\ 15. l-49x 6 . 16. 36 a 4 -!. 

17. 36 a 4 -25. 18. a 4 -49?/ 4 , 19. 121 a 2 - 36 b\ 



FACTORING OF BINOMIALS 77 

20. a 2 6 2 -4c 6 . 21. 49- 100 if. 22. 64a 4 6 4 -c 4 . 

23. a; 4 -144. 24. 9a?-{ a*. 25. 81 a 6 - ^ ?/ 4 . 

26. 5 a b y 5 a?/ 3 . 

SOLUTION. 5 a 5 y 5 ay 3 = 5 a?/(a* ?/ 2 ) 

= 5 ay( 2 -I- y)(a 2 - |/). ^Iws. ( 50, a.) 



27. 3 a 5 -3 ay 2 . 28. 4a 2 c-9c 5 . 29. 20a 3 6 3 -5a&. 

30. 5x 6 5. 31. Sar 5 Sa^ 2 . 32. 6m 4 ?/ -6m 2 ?/. 

33. 8z 10 -8/. 34. 15a 5 -15a& 4 . 35. 32 ax 4 - 32 a?/ 10 . 

36. Make a rule for factoring a binomial that is the difference 
of two squares. 

(1) If a monomial factor appears, what is done? 

(2) The difference of the squares of two quantities equals ... 

37. a 2 m 2 -m 2 n*. 38. aW-a&c. 39. 2mV-8mV. 
40. 3a 5 -12a 3 c 2 . 41. 4a 2 c-9c 3 . 42. 16 a 17 -9 



II. Binomials that are the Difference of Two Cubes. 
Type form : a 3 - 6 3 = (a - b)(a 2 + ab + 6 2 ). 

By 47, Th. VII, the difference of the cubes of two quantities equals 
the difference of the quantities times the sum of the square of the first, 
the product of the first and second, and the square of the second. 

43. w s -n s . 44. p s -q*. 45. (2a) 3 -6 3 . 

46. 8 a 3 -27 ft 3 . 

SOLUTION. 8 a 3 - 27 6 3 = (2 a) 3 - (3 6) 3 

= (2 a - 3 6) [(2 a) 2 + (2 a) (3 5) + (3 6) 2 ]. 
Thus, 8 a 3 - 27 6 3 = (2 a - 3 6) (4 a 2 + 6 a& + 9 6 2 ). 

PROOF. 4 a 2 + 6 a& + 9 6 2 

2a -36 



18 

-12a 2 6 -- 18a6 2 -276 3 
-276 3 



78 FACTORING 

Factor the following and prove by actual multiplication : 
47. 27m 3 -64?i 3 . 48. 125 a 3 -Si/ 3 . 49. 27crV-8. 

50. 2/ 6 -64. 51. 8/-2V 52. y 9 -8. 

53. 216a-6 3 . 54. 343 - or*. 55. xh/ 3 -512z 3 . 

56. a 5 -27 a 2 . (50, a.) 57. 250-2a 3 . 58. 4 a 4 -4 a. 

59. Make a rule for factoring the difference of two cubes. 
How is the first factor found ? How the second ? How is the 
work proved correct ? 

60. m 6 125. 61. m*n mn*. 62. x l5 --6x Q . 

63. 64/-729z 3 . 64. aW-729. 65. 512w w -l. 

III. Binomials that are the Sum of Two Cubes. 
Type form : a 3 + A 3 = ( a + A) (a 2 -- a6 + A 2 ). 

By 47, Th. VIII, the sum of the cubes of two quantities 
equals the sum of the quantities times the square of the first, less 
the product of the two, plus the square of the second. 

Factor the following and prove by actual multiplication : 

66. x 3 + /. 67. Stf + f. 68. m 3 + 27 n*. 

69. 125 + a 3 . 70. aV + 1. 71. 128 + 2 /. 

72. 54 a 4 + 250 a. 73. 3m 5 + 375m 2 . 74. 125 x 6 + 27 y 15 . 

75. Make a rule for factoring the sum of two cubes. What is 
the first factor ? How is the second found ? How is the work 
proved correct ? 

76. 64 a 9 + 125 6 6 . 77. 27m 3 + 512w 3 . 78. 729 x 12 + 64 ?/ 15 . 
79. 8 x 18 + 27 y 12 . 80. 8 3 ?/ 6 s 9 + l. 81. 



IV. Binomials One or More of Whose Factors can be Factored. 

82. Separate a 6 b 6 into its prime factors. 

SOLUTION, a 6 -- 6 6 =(a* + & 3 )( 3 -- ft 3 ). (By I p. 76.) 

= (a + 6) (a 2 - ab + 6 2 )(a - 6) (a 2 + a& 4- & 2 ). 

(By II and III.) 



FACTORING OF TRINOMIALS 79 

83. m* n*. 84. x G --l. 85. 64 a 6 -/. 

86. a? 4 -81. 87. 16 a 4 -!. 88. x-y*. 

89. 16 a 4 -- 81 y\ 90. a?y 625. 91. a 6 ?/ G -z 6 . 

TRINOMIALS 

52. The Factoring of Trinomials. 

L Trinomials that are the Product of Two Binomials having a 

Common Term. 

Type form: jr+(a + 6) JT + ab = (x+a} (x + 6). PROOF 

1. Factor ic 2 + 12 # + 32 and prove the answer. x + 4 

x + 8 
SOLUTION. Two numbers must be found whose sum is 12 ^ _^ 4a . 

and product 32. (See 45, Th, IV.) These numbers are + 8x + 32 
4 and 8. Thus x 2 + 12 x + 32 = (x + 4) (x + 8) . ^Iws. x - 2 + 12 x + 32 

Factor the following and prove by multiplication : 
2. a^+6 +8. 3. ^ + 80; + 15. 

1. 2 +16a+48. 5. 

6. ^ 2 + 8ic + 16. 7. 

8. m 2 + 10 mn + 25 % 2 . 9. or?/ 2 + 22 a^ + 121 z 2 . 

10. 0^ + 26^ + 1690;. (50, a.) 11. a^ + a + J. 

-^2 x 2 10 a/* + 21 SUGGESTION. Sum = 10 ; product + 21. 

Ans. (x 3)(x 7). Prove by multiplication 

13. a 2 -11 a; + 18. 14. aV - 21 a.r; + 80. 

15. or 9 -25^ + 1502/ 2 . - 16. m 2 - 28m + 196. 

17. aj" - 13 ay + 12 /. 18. aV - 21 aa? + 90. 

19. a 3 - 6 a 2 6 + 9 a6 2 . 20. 5a 4 -10a 2 6 + 56 2 

21. a^> - 12 xW + 27 .^6 3 . 22. z 4 - 26 x 2 y + 48 f, 

23. o^ + iC 30. SUGGESTION. 6 +( 6)= 1 ; 6 x - 5 = - 30. 



80 FACTORING 

24. x* + x 6. 25. x 2 z-6. 

26. a 2 + 3z-28. 27. o^-Sa^ 

28. x 2 + 7x-44. 29. m 2 + 6m--16. 

30. z 2 + 5z?/-36?/ 2 . 31. a 2 + 13 a -48. 

32. 2a,- 3 -22x 2 -120a. 33. 3 x 8 - 51 aty + 48 

34. 3x 6 39z 3 + 66. 35. 5 ar> + 30 a 2 - 35 



36. Make a rule for factoring trinomials into binomials having 
a common term. 

(1) What should one look for first ? (See 50, a.) 

(2) How is the common term found ? Ans. By extracting the 

square root of a term in the given trinomial. 

(3) What two conditions must the other terms fulfill ? 

(4) How is the answer proved correct ? 

37. x* - 11 mx + 30m 2 . 38. x*z* + 12 xz 13. 
39. 2z 8 -4o: 4 -48. 40. 35 + 2^-a; 2 . 

41. z 8 - 16 a 4 + 55. 42. mV - 20 wra + 99. 

43. 2xy 6 -10tfy s -28o?. 44. a?x* -7 

45. a 2 -ay-210y 2 . 46. a 2 2 - 38 aa + 361. 

47. m 2 + |m--J. 48. 

49. c 2 d 2 -|cde-- 3 /e 2 . 50. or - 23 xy + 132 /. 

II. Trinomial Squares. 
Type forms: 



51. 

SOLUTION. Vx^ = x ; V<54 = 8. Ans. (x + 8) 2 . 
CHECK, (x + 8) 2 = x 2 + 16 x + 64 by Th. I, 45. 



FACTORING OF TRINOMIALS 81 

52. x* + Wxy + 25y*. 53. 9 z 4 + 24 z?y 2 + 16 y 4 . 

54. 36 or 2 + 60^ + 25. 55. 64 x 2 - 48 xy + 9 y 2 . 

56. 0^ + 40? + 4. 57. (rlr + 2 a&cd + c 2 d 2 . 

58. 9/-30/ + 25. 59. 36 a 4 - 84 ofy + 49 /. 

60. x 2 -x + i. 61. A/-- 1 ^ 2/4-25. 

62. 5a 4 -10a 2 & + 5& 2 . 63. 24 to 3 - 72 bW + 54 & 3 a. 

64. Make a rule for factoring trinomial squares. 

(1) What is to be looked for first? (See 50, a.) 

(2) How is the first part of the answer found ? The last part? 

(3) How is the result checked ? (See 45, Ths. I and II.) 

65. 12 x 2 --36 a? + 27. 66. a&c 10aW+25aW. 

67. 144ar + 600a^ + 625?/ 2 . 68. 49 a 2 + 126 vy 2 + 81 if. 

69. 49 n 16 + 84 m s n 8 + 36 m 8 . 70. 25 a w & 10 - 20 a 5 6VcZ 2 + 4 c*d*. 

71. 18 a 4 + 60 o?b + 50 a 2 6 2 . 72. 121 a 2 6 2 + 176a6 3 + 64 6 4 . 



III. Trinomials that are the Product of Any Two Binomials. 

Type form : ax 2 -f 6jr/ + c/ 2 = (mjr + ny)(px -f y/) 

= mpx 1 4- (my + /?/?)jr/ + fl?/ 9 . 

73. Factor 6 x* - 25 .T?/ + 4 ?/ 2 . 

SOLUTION. Such problems are usually solved by trying sets of factors 
until a pair is found which multiplied produce the given trinomial. We will 
call this the "trial method." It is easy to select two binomial factors for 
trial multiplication which will give the first and last terms of the given 
trinomial. Thus, 

2x y 2x 4y Qx y 

3 x 4y Bx y x -4y 



2 - Bxy 
- 8xy + 4y 2 - 2xy + 4y 2 - 24 xy + 4 



6 x 2 - 1 1 xy + 4 */2 6 x 2 - 14 xy + 4 y* 6 x 2 - 25 zy + 4 f 



82 FACTORING 

Each set gives 6 x 2 and 4 ?/ 2 , but only the last pair of factors gives the re- 
quired middle term, 25 xy. Hence, 

6 x 2 - 25 xy + 4 y 2 - (6 x y) (x - 4 y). 

To factor by this method try different pairs of factors until a combination 
of signs and coefficients is found which produces the given quantity. The 
middle terms are called " cross products " ; the others, end products, 



74. 9ar + 9z?/ + 22/. 75. 2x? 

76. 2x- + 5xy + 2y 2 . 77. 3 x 2 + 13 xy + 12 

78. 3 or- 13 x + 14. 79. 6 a? -31 a; + 35. 

80. 8x?-3Sxy + 35y 2 . 81. 4m 

82. 3x- + 7x-6. 83. 3 m 2 - 19 mn - 14 w 2 . 

a. If a given trinomial does not contain a monomial factor, evidently wo 
factor of it can contain that quantity, and there is no use trying factors which 
contain a monomial divisor. Thus, in solving Ex. 73, it was a waste of time 
to try the second set of factors, since the first of these, 2 x 4 y, contained 
the factor 2, while the given quantity did not. 



84. 6z 19x + 15. 85. 

86. 12x 2 + 7x-l2. 87. 



*6. In the type form mp 2 + (mq + np)xy + ftQ'?/ 2 given above, observe that 
mq + np is the swm of two quantities whose product is mp x wg. We look 
first for two numbers whose sum is the given middle coefficient and whose 
product is the product of the given first and last coefficients. 

Thus, in factoring 4 x 2 -- 73 x + 18, 

(- 72) + (- 1) = - 73, and -- 72 x - 1 = 4 x 18. 

We choose our coefficients now so that one cross product will be 72, and 
the other 1. Thus, 

4x - 1 
* Seel, p. 6, x -18 

4z 2 - x 

-72 3: + 18 

4x 2 -73x+ 18 



FACTORING OF QUADRINOMIALS 83 

88. 3x?-5x~2. 89. 4 ar 8 - o;y 3 # 2 . 

90. 4ar J + 16a? + 15. 91. 24a 2 - 29 a*/- 

92. 8 a? 4 --19 or 8 15. 93. 40 + 2x-2ar } . 

94. 36 it- 4 - 18 x 2 - 10. 95. -10a 4 + 7 

96. 40 + 6 a - 27 a?. 97. 16 a 5 + 4 ofy 2 - 30 

98. Make a rule for factoring trinomials that are the product of 
any two binomials. 

(1) How are the coefficients of the first terms of each binomial 

factor chosen ? (See Ex. 73, p. 81.) Those of the last terms ? 

(2) When a set of coefficients and signs are chosen, what is done ? 

How can selection of coefficients be guided ? (See a, 6, p. 82.) 

99. 12 x 2 - 31 x -15. 100. 24 + 37 a; -72 a 2 . 
101. 20-9c-20c 2 . 102. 24x* 35xy + 4:y*. 
103. 8 m 4 + 38 mV -f 35 n*. 104. 132a 2 + a-l. 
105. 12^ + 50^-50. 106. 3 a 6 + 41 a 3 + 26. 
107. 15 a? 8 + 224 a; -15. 108. loa 8 - 77 a 4 6+106*. 

QUADRINOMIALS 

53. The Factoring of Quadrinomials. (Polynomials of four terms.) 

I. Quadrinomials that are the Cube of Binomials. 
Type form : a 3 - 3 a 2 6 + 3 ab 2 - 6 3 = (a - 6) 3 . 

1 . 8 a 8 - 60 afy + 150 xy 2 - 125 y\ 

SOLUTION. We extract the cAtbe root of the first and last terms, getting 2 x 
and 5 y. Then (2 x 5 y) 3 is the answer. 

PROOF. 

(2x - 5 y)3 = (2 x) 3 - 3(2 z)2(5 y)+ 3(2 x)(5 y)2 -(5 y) 3 . ( 45, Th. V.) 
= 8 x 3 - 60 x 2 y + 150 z?/ 2 - 125 y s . ( 27.) 

2. 646 3 + 486 2 + 126 + l. 3. 216 x 3 - 108 x*y + 18 z/ - 

4. 8 a 3 - 36 a*& + 54 a& 2 - 27 6 3 . 

5. 27 m 9 + 108 mV + 144 m 8 w 4 + 64 n*. 



84 FACTORING 

II. Quadrinomials that are the Product of Two Binomials. 
Type form : ac -f- ad + be + bd = (a + b) (c + </). 

6. Factor 5 ax ex 5 ay + cy, and prove answer. 

SOLUTION. It is helpful to recognize three steps in the solution. 

(1) Putting two terms in one parenthesis and the remaining two in another, 

(2) Removing monomial factors from these quantities in parenthesis. 

(3) Dividing by the common factor as in 50, Ex. 29. 

1st step (5 ax ex) (5 ay cy}. ( 33.) 

2d step (5 a c}x (5 a c)y. (Taking monomials out, 50.) 

3d step 5 a c\ (5 a c)% (6 a c).y (Short division.) 

x - y ( 50, Ex. 29.) 

4th step. PROOF. 5 a c 

x-y 

5 ax ex 
5 ay -f cy 



5 ax ex 5 ay + cy 

Factor and prove in the following : 

7. a 3 -a*c + ac 2 c 3 . 8. 3x* - x* + 6 x- 2. 

9. 6a s -6a 2 y + 2ay*-2y\ 10. 3 c 4 - 3 c s n + en 2 - n s . 

11. x 5 + # 4 + 4 # -f 4. 12. ab by a + y. 

13. 20 ad 35bd 8 ax -f 14 bx. 14. x* + 3y -3 a xy. 

15. a 2 x 2 6 2 x 2 ay + 6 2 z/ 2 . 16. 3a?y+3aby 5 any 5bny. 

17. 4a 2 -aV + a 2 -4. 18. 4aar 5 + 8 a.T- 8 a -lax 2 . 

19. Make a rule for factoring quadrinomials that are the prod- 
uct of two binomials. 

(1) What is the first step? the second? the third? 

(2) How is the answer proved correct? 

20. 3x 3 -6ax' 2 -x + 2a. 21. 2 xy + 3 yz + 6 y* + xz. 

22. 4?/ + 9z 3 -6rV-6#. 23. 6 a 2 + 9c- 6&c-4a&. 

24. 12 fta? + 3 ay 9 6i/ -- 4 aaj. 25. 6 a 2 - 3a&-- 26m -f 4am. 

26. amx 2 + bmxy - anxy bny 2 . 27. / V + g-x- ag* of 2 . 

28. x 4 -4ar + 3ar>z 2 -12z 2 . 29. 



FACTORING OF QUADRINOMIALS 85 

III. Qiiadriuoniials, the Difference of Two Squares. 
Type forms : 
(1) x- + 2 xy +/ - z 2 = (x +/) 2 - z 2 

= [(*+/) + <][(*+/)-*] (% 51, I.) 

= [a+(6-c)][fl-(6-c)]. (By 51, I.) 

30. Factor # 2 6 ax 16 b 2 + 9 a 2 and prove result correct. 

1st step (x 2 6 ax + 9 a 2 ) 16 6' 3 . (On putting three terms which consti- 
tute a square in parenthesis. ) 

2d step (x 3 a) 2 (4 6) 2 . (On changing to form of squares, 52, II.) 
3d step [ (x 3 a) + 4 6] [ (x 3 a) - 4 6] . (On factoring, by 51 , I. ) 
4th step [x 3 a + 4 6] [x 3 a 4 6]. (On removing parentheses, 32.) 

5th step. PROOF, x 3 a -f 4 & 

a- 3a -46 
# 2 3 ax + 4 &x 

- 3 ax + 9 a 2 - 12 a& 



a 2 - 6 ax + 9 a 2 - 16 

Solve the following as the model and prove : 



31. a a -4a6 + 46 2 -9c 2 . 32. ar- 

33. a 2 + ?/ 2 + 2a,'?/-4ary 5 . 34. a 2 

35. 9 2 -a 2 -6a +1. 36. 4ar J 



37. a 2 - 36 6V 9 + 126a-l. 

SUGGESTION 



Finish solution as in Ex. 30. 

38. a 2 + 12 ex 9 y? 4 c 2 . 39. 4c 2 a 4 9& 2 6a 2 &. 

40. 9a 2 -a 2 -4a&-4& 2 . 41. 36 & 2 -4 + 20 a -25 a 2 . 

42. Make a rule for factoring quadrinomials that are the differ- 
ence of two squares. Describe five steps. 



86 FACTORING 

43. 25# 2 -4a 2 + 162/ 2 +40a2/. 44. 49# 2 + 4?/ 2 - 25z 4 -28z?/. 

45. (x 5 ?/) 2 4 6 2 . SUGGESTION. Only three steps remain here. 

46. (3o;-22/) 2 -36m 4 . 47. 9 x 2 - (4 y - 1 z) 2 . 

OUTLINE FOR FACTORING 

54. General Methods in Factoring. Leaving that case in factor- 
ing in which, monomial factors are removed from polynomials 
( 50) out of consideration, we can say that all the problems of 
this chapter can be factored in one or other of seven ways : 

1st. If they are binomials, by one or other of the three Theo- 
rems III, VII, VIII ( 45, 47). 

2d. If they are. trinomials, by the trial method. 

3d. If they are quadrinomials, by Theorem V, or by arranging 
their terms as follows : either (a) By putting two terms 
in one parenthesis and the remaining two terms in another 
parenthesis, and then removing monomials ; or, (5) By 
putting three terms in a parenthesis to form the square 
of a binomial. 

a. More difficult cases in factoring are given in a later chapter, and tri- 
nomials will be found there which cannot be factored by the " trial " method, 
as also binomials, that cannot be solved directly by the theorems mentioned 
above. It should be noted here that while such a trinomial as 9 a 2 + 30 ab + 
25 b 2 can be solved by the trial method, it can be solved more easily by observ- 
ing that it is the square of a binomial. 

55. Miscellaneous Exercise in Factoring. The student should 
carefully memorize the methods of solution for the different kinds 
of problems, especially the seven cases of the preceding article. 
In case of trouble, note whether the given quantity is a binomial, 
a trinomial, or a quadrinomial. Then study to find which case, 
in the article in which it falls, it comes under. Always look 
first for a monomial factor, and if any is found put it before or 
after a parenthesis containing the other factor. 



MISCELLANEOUS EXERCISE IN FACTORING 87 

Separate the following quantities into their prime factors, and 
by multiplying the factors together, unless the factors are 
actually multiplied together in the solution itself : 

1. x- + llxy + 30y' 2 . 2. a 2 + 16 a; + 63. 

3. a 2 -25. 4. 36^-60^ + 25. 

5. x 7 + x s y + x*. 6. 4 x 2 + 4 xy + y\ 

7. 36 ^-9/. 8. x 2 22 a; + 121. 

9. a 4 --ra 4 . 10. 

11. crb 4 a 4 b 2 . 12. 

13. a 3 + 27. 14. 

15. 16 -a 4 . 16. 

17. 125 a 3 + 27. 18. 

19. x^ Sy 3 . 20. 

21. a; 2 -a -2. 22. 

23. 125 /- 27 a*/ 3 . 24. 6 + 15 a 2 -19 a. 

25. ar 3 -4^-45aj. 26. 169 a 2 + 169 b~ - 338 ab. 

27. x 4 - 16 or 2 + 55. 28. 
29. mV + 1. ( 47, VIII.) 30. 

31. a 6 6 6 . 32. 

33. 2m 2 /i--4mw + 2n. 34. 

35. x w - 26 a 5 + 168. 36. 

37. 27 a 3 + 64 6 3 . 38. aV 5 a 3 # - - 14. 

39. 1 x lG . 40. 

41. ar 5 x. 42. 

43. 0^-49 x + 48. 44. 5 x> + 30 ofy 2 - 35 

45. # 2 ax + <./ ac. 46. 

47. 250 a;- 2 x 7 . 48. 



88 FACTORING 

49. 40a 2 +61a6-846 2 . 50. 56 aa?y + 96 axy + 

51. a 8 -1692/ 10 . 52. 3 a 8 -51 a 4 + 48. 

53. 5 ax 9 5 a. 54. 2 x 4 2 x 3 2 a-x 2 + 

55. 110-x-x 2 . 56. 24^ + 5 xy-36y 2 . 

57. a + 343 a 4 . 58. x 5 x 4 -- 16 x + 16. 

59. 21 a 2 17 a 30. 60. ax 2 - - ex + ax c. 

63. 4a 2 + 42/(j/ 2x). 64. 64 x 6 729 ?/ 6 . 

65. 8^ 3 +27P 3 . 66. a 3 --a* + GM 2 - 

67. c 2 -2c-399. 68. 243 6 5 c 6 - 75 6 7 . 

69. x l5 y }5 . 70. 

71. M 6 -N 6 . *72. 22 a 2 - 110 f - 

i ^J C( " I "/j~ \AJ ~^p* I * ^v ^ C/ ^^^ tJ vv C/ I Cv ^"^^ 

O O * 

75. a 9 6 9 . 76. acx 2 + adx -{- box + 

77. P 2 -P-72. 78. 49s 4 + 70s + 25 

79. a 8 -^-ft 3 . 80. 

81. 121 ^ 12 - 324 7i 18 . 82. a; 2 - a + y2 - a;^. 

83. a; 6 + 64 & 6 . 84. 1 a 2 + 2 a& b 2 . 

OO W W />" "" *v ~" J- OO (l/ \jJU ""l " tl' CvA/ " ~" L/C'*// " ~ C/C*e 

87. x 12 + i/ 12 . 88. a 2 - 

QO r'"ir^ /v^ojiO Or^ '>*-' 

Ot7. J- *~~ ^tJ\J *Hs U * *J\J* *O 

Q1 ^/'T /Tf"'V* I ft nw" OO r /v?*6 

</ X . ^ i C( *<.- ' ~ J_O C(/*X/ \J ^i . - **/V 

93. 343 t 6 --u 3 . 94. a 6 5 # ap +p.r. 

95. 400 n 16 - 484 ^ 16 . 96. 12 a^/ - 28 ajy 3 - 9a 2 kr 2 +21 

97. 88m 9 -297?i 6 . 98. 8.v 3 + 12 a 2 + 6 a; + 1. 

99. 9 a; 2 -82 a? + 9. 100. 3- 



PRIME QUANTITIES 89 

56. Prime Quantities. It will be found that many quantities 
called prime in this chapter can be factored by using fractional 
exponents. Such quantities are prime in the sense that no quan- 
tities with integral exponents other than themselves and unity 
multiplied together will produce them. 

All the quantities in the preceding articles were factorable. 
But quantities taken at random are more likely to be prime than 
composite. The student should learn to know when quantities 
are prime, as well as how to factor them when they are composite, 
since then he will not waste time trying to factor them. 

1. TJie sum of two squares is prime (unless it is also the sum of 
odd powers of other quantities). Thus, a?-\-b 2 is prime, but the 
binomial a 6 + 6 6 , which can be regarded as (a 3 ) 2 + (6 3 ) 2 , is not 
prime, since it can also be written (a 2 ) 3 -f (6 s ) 3 , or as the sum of 
two cubes, and as such can be factored. 

2. The expressions x 2 + xy -j- 2/ 2 and y? xy -f- y 2 found by di- 
viding a? y* by x y and a? 3 -f y 3 by x + y are prime. 

3. Expressions like a 3 + b 2 , a*--b 3 , etc., in which the powers 
are not the same and cannot be viewed as the same without 
making the exponents fractional, are prime. Notice that such 
an expression as a 12 -f b 9 can be regarded as the sum of the same 
powers by writing it (a 4 ) 3 -f (6 3 ) 3 . 

4. Can z 4 + ?/ 4 be factored ? Why ? Can x 8 + y 8 ? Why ? 

5. Can 4 x 2 -f 6 xy -f- 9 y- be factored ? Try it. See 2 above. 

6. Can 25 x 2 30 xy + 9 y 2 be factored ? 

7. Can 25 x 2 - - 15 x + 9 be factored ? 

8. Can m 15 + n 9 be factored ? m 15 + n 14 ? 

9. Can x- 1 - y 15 be factored ? x 21 - y w ? 

10. Can 6 x 2 + 7 x -f 11 be factored ? Try all combinations. 

11. If the coefficients of a trinomial, a binomial, or a quadri- 
nomial are selected at random, is it likely that the resulting 
quantity can be factored ? Try it, and find out. 



90 FACTORING 

QUADRATIC EQUATIONS BY FACTORING 

57. Solution of Quadratic Equations by Factoring. The degree 
of an integral equation equals the exponent of the highest power 
of its unknown. Equations that contain the first power of the 
unknown, as x, and no higher power, are called simple equations, 
or equations of the first degree. Equations that contain the 
second power of the unknown, as x*, and no higher power, are 
called quadratic equations, or equations of the second degree. 

1. Solve the equation x z - 5 x -f- 6 =0, and prove answer correct. 
SOLUTION, x 2 5 x + 6 = (x 3) (x 2) = 0. (On factoring, 52.) 

Each of the two. factors is now set equal to zero. If a product is zero, a 
factor of it must equal zero. 

x 3 = 0, whence, on transposing, x = 3. Ans. 
x 2 = 0, whence, on transposing, x = 2. Ans. 



PROOF. If x = 3, x 2 5 x + 6 = 
becomes 3 2 5x3+6=0 

or, = 



If z=2, z 2 5 z + 6 = 
becomes 2 2 5 x 2 + 6 = 
or, = 



a. To see why the process just explained solves the problem, notice that 
both x = 3 and x = 2 substituted in the factored form of the given equation 
make the two sides equal. 



Thus, if x = 3, (x - 3) (x - 2) = 
becomes (3 2) = 

or, = 



If jc = 2, (z-3)(B-2)=0 
becomes (2 -- 3) = 

or, 0=0 



( 35, 21, 22.) 

Solve and prove both answers in the following. The seconcl 
member must be made 0, if it is not already so, before factoring : 



2. z 2 -8a+15 = 0. 3. x 2 - 

4. ar +-.8^ + 15 = 0. 5. i 2 + 9ic + 20 = 

6 . x*-2x-15=0. 7. a? -7 a; = 18. 

8. a 2 -12 a -85 = 0. 9. o 2 -19. = 20. 

10. 3 x 2 11 x + 6 = 0. SUGGESTION. Factor as in 52, III. 



SOLUTION OF QUADRATIC EQUATIONS BY FACTORING 91 

11. 2 x 2 + x -15 = 0. 12. 6 a 2 -11 a- 35= 0. 

13. 15a>*-14a;-8 = 0. 14. IS a- 2 - 3 a = 36. 

15. 49^-35^=66. 16. 40-17z-5r J = 0. 

17. Make a rule for solving quadratic equations like the pre- 
ceding by factoring. 

(1) What must the second member be ? 

(2) What is done with the left member ? 

(3) What is the next step? 

(4) How many answers are there to prove ? 

18. 3x* + 5x = 2. 19. 
20. 30* + 10 a; = 57. 21. 
22. 4ar = 3z + 27. 23. 

24. 6 a; 8 -96 = 0. 25. 3x' 2 -12 = B 2 

26. 5a 2 + 19 = 12a 2 -9. 27. 7x 2 - 13 = 2o; 2 -8. 

28. Find a number such that its square is equal to 72 more than 
the number itself. 

29. Find a number such that if 17 times the number is di- 
minished by its square, the remainder will be 70. 

30. Find a number such that if you subtract it from 10 and 
multiply the remainder by the number itself, the product will be 21. 

31. What two numbers are those whose difference is 2 and the 
sum of whose squares is 130 ? 

32. Find three consecutive numbers such that the sum of their 
squares equals 110. 

SUGGESTION. Let x = first no. ; x + 1 = second ; x + 2 = third. 

33. The average height of men who weigh 120 Ib. is a certain 
number of feet. Three times the square of this number increased 
by 9 times this number equals 120. What is the average height 
of 120 Ib. men ? 

b. In such problems as this, the negative answer has no meaning and is 
disregarded. 



92 



LOWEST COMMON MULTIPLE 




33 



23X 



34. The perimeter of a room, that is, the distance around it, is 
30 ft., and its area is 54 sq. ft. Find its width. 

SUGGESTION. Let x = the number of feet in the width. Then 15 x = the 
X length, and (15 x)x = 54. Why ? 

35. A field is 35 rods long and 
23 rods wide. It is desired to 
border it all around with a strip 
of the same width to contain 11 
acres. How many rods wide 
shall this strip be ? 

36. A farmer cuts grain in a 
field 100 rods bv 80 rods. How 

tt 

many rods wide must the strip around the field inside the fence 
be to contain 4.4 acres ? 

37. The number of different calls possible in a small telephone 
exchange having x subscribers is x(x 1). If the number of calls 
is 110, how many subscribers are there ? 

38. What is the radius r of the base of a cone whose slant 
height is 11 ft. and total area 660 sq. ft. if it can be found by the 
formula ?- 2 + 11 r = ^ X 660 ? 



LOWEST COMMON MULTIPLE 

58. Lowest Common Multiple of Quantities. 

1. A multiple of a quantity is the quantity multiplied by some 
integral quantity. Hence a multiple of a quantity is divisible by 
it. Thus, 6 a 3 is a multiple of 2 a. 

2. A common multiple of two or more integral quantities is an 
integral quantity that is exactly divisible by each of them. The 
word " common 5: signifies "belonging to each or many." A 
common multiple of two or more quantities belongs to each in 
the sense that it contains each quantity. 

Thus, 36 4 & 3 c 2 is a common multiple of 3 a 2 b, 9 abc, and 2 abr, 



LOWEST COMMON MULTIPLE 93 

3. The lowest common multiple of two or more integral quanti- 
ties is that quantity of lowest degree which is exactly divisible 
by each of the given quantities ; its coefficient (Arabic notation) 
is the least number that will contain all the coefficients of the 
different quantities. 

Thus, IScrfrc 2 is lowest common multiple of Sorb, 9 abc, 2 abc 2 . 

4. Find the lowest common multiple of a 2 b, 3am, 9m 2 . 

Ans. 9 arbm 2 , 
PROOF. 9 a 2 &m 2 -s- a-b = 9 ra 2 ; 9 2 6m 2 -4- 3 am = 3 abm ; 9 2 &m 2 -=- 9 m 2 = a?b. 

Moreover, if any quantity with a smaller coefficient or with lower ex- 
ponents of the letters had been taken it would not have contained all of 
the given quantities. 

Find the lowest common multiple in the following exercises and 
prove the answers correct : 

5. 3 a-b, 2 ab 2 , 4 a 3 6c. Ans. 12 aVc. 

6. 4a, 2a 2 b 2 , 12 a 3 . 7. 6x>y,Sy*z. 
8. 12 a 2 x 2 , 9 a 2 /, 4. 9. 2 a, 3 b, c. 

10. 8 a V, 30 aV, 4 aV. 11. 3a?b,2ab*. 

12. 14 a 3 , 21 a 2 , 5 6, 7 a. 13. 7 a s m 2 , 21 wV. 

14. 42 a 3 &* and 60 c<?b\ 15. 24 aV, 60 aV. 



16. 

SOLUTION. 3 a& + 3 6 2 = 3 6 (a + 6). ( 50.) 

a 2 + 2 & + 5 2 = (a + fr) 2 . ( 52, II.) 

L. c. m. = 3 &(a + 6) 2 . (Definition in 3 above.) 
3 6 (a + 6) 2 -f- 3 b(a + 5) = a + b ; 3 6(a + 6) 2 ^ (a + 6) 2 = 3 6. 

In the following, first factor each composite quantity and then 
write the 1. c. m. by 3 above. Then test answer. 

17. 6m 2 6 mn, 3 m 4 3m 3 n. 18. a + 1, a? - 1. 

19. 2x + 2,x*2x 3. 20. x + 1, x 1. 

21. a 2 5 a;, 0^3 x 10. 22. or 3 re, or 5 --!. 

23. or 5 a 3 , 3 y? + 3 ax + 3 a 2 . 24. m 3 + n 3 , m + n. 



94 LOWEST COMMON MULTIPLE 

25. Make a rule for finding the 1. c. m. of quantities 

(1) What is done first with each quantity and each coefficient ? 

(2) How is the coefficient of the 1. c.m. found? Ans. As in 
arithmetic. Thus to find the 1. c. m. of 8, 4, 9, 18, 14, we cancel 4 
and 9, since 4 is contained in 8, and 9 in 18, and any number th&t 
will contain them will contain 4 and 9. Then 8 = 2 3 , 18 = 2 3 2 , 
14 = 2 7. Taking each prime factor with its highest exponent, 
we have 1. c. m. = 2 s 3 2 7 = 504. 

(3) What exponent of each literal part factor is taken ? 

(4) What is done with the coefficient and literal factors found ? 

26. 9 ar, 15 xy t 18 xy s . 27. 15 mn 2 , 54 m/i 3 , 90 m 2 n. 

28. 24(a + 1), 32(a - 1), 60 a. 29. a?-y*, a?-3xy + 2y*. 

30. 3 a- 3 -3 a, 6a?-12x + 6. 31. x 2 - 3 a? -40, 3a 2 + 15 *. 

32. cr?> + a& 2 , arb-ab*, 3a 2 -36 2 33. ax%T- ?/) 2 , bxy(x 2 - y*\ 

34. a; 2 3 a; -54, or-lOa + 9. 35. o? y 2 ,a? 8xy + 7y 2 . 

36. a 2 - 6 2 , a 3 - & 3 , a 3 + 6 s . 37. a 2 - f, (x - ?/) 2 , a 3 - /. 

38. a? 1, or 2 -!, a? 2, 2 -4. 39. a; 2 ?/ 2 - 4, a 2 - 8 xy + 12. 

40. 2^ + ^-1,4^-1,2^4-3^ + 1. 

41. a; a, a 2 a? 2 , ic 4 a 4 . Ans. x* a 4 . 

a. Since the lowest common multiple deals solely with the question )1 
divisibility, in getting the 1. c. m. any given quantity can have its sign changed 
by multiplying it by 1. In the example just given, a 2 oc 2 can be changed 
to x 2 a 2 . The 1. c. m. x 4 a 4 contains a 2 x 2 without remainder 

42. a 3 x s , a 2 2 , x a. 

43. 1-205,4^-1,4^ + 1. 

44. a? + x 12, -36 + 13 07-a^, a 2 -16. 

45. 6 2 -a 2 , 46 + 4a, a 3 - 3 a 2 6 + 3a6 2 - 6 s . 

46. 1 - - x, x 2 1, a 2, 4 or*. 

*47. 4(1 - a) 2 , 8(1 - a), 8(1 + ), 4(0* - 1). 



LOWEST COMMON MULTIPLE 95 

48. 3ar'-5z + 2, 4or J -4ar J -a;+l. (53,11.) 

49. ar 1 - ar - 9 x + 9, or + 2 x - 3. 

50. a,- 2 - 4 a 2 , ar 5 + 2 ax 2 + 4 a 2 x + 8 a 3 . 

51. x 4 - - y 2 , or 3 + x s y -f- xy + ?/ 2 . 

52. 2x*-xy-15y*, tf-Gxy + Qy*. 

53. or 5 + /, a; + y, xy s? - - if. 

54. (x - l)(a? + 3) 2 , (a; + l) 2 (ar - 3), or - 1. 

55. (a? - 1) 3 , 7 aj/Caj 2 - I) 2 , 14 a*y(x + I) 3 . 



Solve the following problems mentally, giving the answer in the 
factored form : 

56. 3 a 2 6, 4 ac 2 ,. 6 & 2 c. 57. 7 x 2 , 2 aj 2 - 6 x. 

58. + &, a- + 2 ab -f 6 2 . 59. 6 a6 2 cd 5 , 9 a W, 15 a6 2 c 3 . 

60. a 2 + a6, 6 - 6 2 , a 2 -Z> 2 . 61. a; 2 -4a; + 3, a,- 2 -5 a; + 6. 

62. x* 4- 8 a; + 15, x 2 + 4 x 5. 63. a? 5 x 84, a; 2 7 a; 60. 

64. 6 db(a + b) 2 , 4 cr(a 2 - 6 2 ). 65. a 2 - 9 a? -10, a; 2 - 7 a^-30. 



CHAPTER III 
FRACTIONS. HIGHEST COMMON FACTOR 

59. A fraction in algebra is an expressed division. The divi- 
dend, or numerator, is written above a line and the divisor, or 
denominator, below it. The numerator and denominator, as in 
arithmetic, are called the terms of the fraction. The operations 
with fractions are much like those in arithmetic. 

60. Reduction of Fractions to their Lowest Terms. This opera- 
tion depends on the well-known principle explained in arith- 
metic, that both terms of a fraction may be multiplied or divided by 
the same number without altering its value. 

To reduce a fraction to its lowest terms, factor both numerator 
and denominator into their prime factors and cancel all factors 
common to both. Evidently canceling common factors is equiva- 
lent to dividing both numerator and denominator by the factors 
canceled. 

Type form : 



1. 



be b 

15aW t 

25 ab 4 c ' 



/ 25 



SOLUTION. ZaWc Y- -^ = ^^-. Ans. 

25 a6 4 c 5 6 

CHECK. Let a = 2, b = 2, e = 1. Then given fraction equals f , and 
answer equals f . 

. Checking here does not show that an answer is in its lowest terms. It 
merely shows that the answer has the same numerical value as the given 
quantity 

96 



REDUCTION OF FRACTIONS TO THEIR LOWEST TERMS 97 

Reduce the following to their lowest terms and check several 
problems : 

108 a s b 2 30 oV SaV 110 mx 2 

2< 111 m' 3 ' o,. J ' 4 ' "TTJ' 5 ' K~n "2* 

144 acr ^ 06 an/ a c or ooO ??I-OT 

45 a W 17a 7 6 5 36 a; 2 y 4 V" 4 * 5 

' 195 aV' ' 19a r 6 6 ' ' 60 x*y 2 ' ' arx 7 

- A y? 7 icy + 10 y 2 (x^2rjfi (x - 5 ) x 5 

10. -/ SOLUTION. -^-^ = - 

ar 8ic?/ + 12?/ : ' (%^y)(x> ti y) x 6 

CHECK. Let x = 3, y =2. 

Then. a; 2 - 7 yy + 10 y 2 _ 7 a i so x - 5 y - ^L _ 7. 

x a _ s a:y + 12 y 2 9 x - y 9 9 



- 



4 a 2 - - 48 aaj + 128 a 63(a 2 - I*) 



x 2 + 2 fly + ?/ 2 . 16 

5 * 



17 -. is 

} -y' 5 )(x + 2/) 
_ 3 + 3 _ 



x~y x 
6(n 2 -2?i 



- 22 

' ' ' 



23. Make a rule for reducing fractions to their lowest terms. 

45 aV - 105 5 ^ 

33 ab s x - ' 



2 



98 FRACTIONS 



*26. - ' ^ ' SUGGESTION. Square, and add terms in 
a " the numerator first. One cannot cancel as the 

problem stands, since neither a + b nor a b is a factor of the numerator. 

a. The student will avoid errors if he will simplify numerator and denomi- 
nator (if necessary), then factor, then cancel common factors. 






24^-22^ + 5 

* ' 



m 2 +7m 



+ 16^-15 ' 9a 2 -9x'-28 

a 2- 

' 



61. Highest Common Factor. 

1. A common factor of two or more integral quantities is a 
quantity that will divide each of them. In the preceding article 
the factors canceled were common factors of both numerator and 

denominator. 

i 

2. The highest common factor of two or more given integral 
quantities is that integral quantity of highest degree and largest 
numerical coefficient which will divide each of them exactly. It 
is the product of all the prime factors that are common. 

3. A rule that can be used in reducing a fraction to its lowest 
terms is : Divide both numerator and denominator by their highest 
common factor. 

4. Find the highest common factor of the numerator and de- 
nominator in each of the exercises of the preceding article. 

5. Find the highest common factor of x* a 4 , a^+a 3 , and x 2 a 2 . 
SOLUTION, x 4 - a 4 =(x 2 + a 2 ) (a; -f a}(x a) 

x s + 3 = (x + ) (X 2 ax + a 2 ) 



H. c. f. = x -f a (By definition in 2 above.) 



SIGNS IN FRACTIONS 99 



6. 

7. l-lla + 18a 2 , 8a 3 -l, 18o-'-oa-2. (See 58, 41.) 
8. 



10. a 2 -3a--18, 2a 2 - a - 21, 3a 2 + 4a- 15. 

11. 2a 2 + 17a-t-36, 4a 2 -4a-99, Ga 2 + 25a-9. 
For additional exercises use the quantities of 58. 

NOTE. Highest common factor and lowest common multiple are com- 
monly put in adjoining chapters. As the processes of finding them are 
somewhat alike, students may confuse the two, particularly if they fail to 
grasp what the terms mean. For this reason highest common factor is here 
treated in connection with reduction of fractions to their lowest terms - 
the only place where this subject finds any application in elementary 
algebra. Indeed, so far as elementary algebra is concerned the student 
can get along very well without h. c. f . by factoring. 

62. Allowable Changes in the Signs of 'the Numerator and Denomi- 
nator of a Fraction. There are three signs connected with every 
fraction : that of the numerator, that of the denominator, and 
that of the quotient which stands before the fraction and is called 
the sign of the fraction. 

Thus ^ ^ has + (understood) for the sign of its nu- 

- (c - d) 

merator, for the sign of its denominator, and for the sign of 
the fraction. 

1. The signs of all the terms of both numerator' and denominator 
of a fraction may be changed, because this amounts to multiplying 
both numerator and denominator by 1, which does not change its 
value. ( 60.) 

Th a + ba b. 2a 3b-)-c2a+3b c 



c d c + d m + r a a--m r 

( 60, 13, 1.) 



100 FRACTIONS 

2. The sign before a fraction and the signs of all the terms of its 
numerator can be changed without altering its value, for this amounts 
to multiplying it by 1 twice, and 1 X 1 = + 1. 

Thus, 



m 4- n m -f- n 

3. The sign before a fraction and the signs of all the terms of its 
denominator may be changed without altering its value, for this 

amounts to multiplying it by -- , or 1, as ( - j = (!)=!. 

~~ 



Thus, 



c b b c 



a. If one or all three of the signs of a fraction are changed, its sign is 
changed ; if any two are changed, its value is not altered. Factors and 
terms of the numerator and denominator should not be confounded. (See 4 
and 5 below.) 

4. The signs of an even number of factors of the numerator, or of 
the denominator, or of both, can be changed without changing the value 
of the fraction, since an even number of Vs multiplied or divided 
give +1. 

Thus (b a)(d~c) = (a - fr)(c - cf) 

m)(q--p) (m n}(p 



5. TJie signs of an odd number of factors of the numerator, or 
denominator, can be changed if the sign before the fraction also is 
changed. (See 2 and 3 above.) 



[ 

c 2 c 2 ' (a? - y) (z - y) (x-y)(y-z) 

b. It is important that the sign of a fraction should always be written on 
the line. Beginners are frequently careless about this, and are led into con- 
fusion as a result. 

Thus, a + - is bad form. It should be written a + - 



c 



ADDITION AND SUBTRACTION OF f FACTIONS U'l* 

63. A mixed quantity is composed of an integral quail tity. and 
a fraction. 

Thus, a and ^ + r are mixed quantities. 
c p + q 

a. It is desirable that the student should note here a difference between 
the Arabic and the algebraic notations. In algebra, when no sign is written 
between quantities, multiplication is understood ; in arithmetic, additicn is 
understood. 

Thus, a - must mean in algebra a times - ; in arithmetic, on the other 
c c 

hand, 3| means the sum of 3 and |- 

I-* 

If one desires to denote the sum of a and _ , he must write + between 

ft c 

them ; thus, a H 

c 

64. Addition and Subtraction of Fractions. 
Type form: = AjB. 



_ a . b A a-\-b 

3. - + - Ans. 2 

c c c 

CHECK. Let a = 2, 6 = 3, c = 6. 

Then 2 + $ = M = 6. also 



c c 66 6 c 6 6 



a b c K a b c d 

4. -H 1 5. 

m m m x x x x 



2x 1 1 + a: . x 2 ru^z, 

6. Ans. - Check. 

1 071 OS 1 



q-f-6 

rr / \ ' p^ 



102 FRACTIONS 



453 4 J b 8 

4 3 3 6 5 25 1 



a; x 4 37 



43 6 

SOLUTION. Fractions in algebra are added by the same process as in 
arithmetic. We first find the lowest common multiple of the denominators, 
calling it the lowest common denominator (I.e. d.). 

Then the 1. c. d. is divided by the denominator of the first fraction and 
both terms of this fraction are multiplied by the quotient. The same is 
done with the other fractions, thus yielding equivalent fractions, since 
multiplying both terms of a fraction by the same quantity changes its form 
but does not alter its value ( 60). Last of all, the numerators are added and 
the sum placed over the common denominator for the answer, which is to be 
reduced to its lowest terms if not already so. 

The 1. c. d. in this problem is 12. 

Thus, 5_ 



43 6 



becomes * - *~ -. * , (60) 
12 12 12 



8s-(4s-16)-.C2s+10) t (As in Exs. 1-8, p. 101.) 

\a 

3% 4x 4- 16 2 .x 10 _ 6 3 x (On removing parentheses 

QjV .j-- __ - - * 

12 12 and adding.) 



But, = n * x > = ?^=-Z. (On reducing to lowest terms.) 

1. *j r~r^ * 

4 

/ 

CHECK. Let re = 4. Tben --^-=^-^-^ = 1 ---? = -!; 

43 6 322 

2 -=-^ = = =--. (62,2.) 

A A A O ^ ** / 



ADDITION AND SUBTRACTION OF FRACTIONS 



103 



a. The student will do well, in solving the problems of this article, to fol- 
lc/w the model pretty closely, beginning solutions by writing down the 1. c. d. 
Later he can shorten the process somewhat by omitting one or more of the 
steps. He should make use freely of checking in fractions, as this is likely 
to be his only means of knowing whether the answer is correct. He should 
be careful to see that his answers are in their lowest terms, as the checking 
does not test for this. If at any time a letter is given such a value that it 
makes any denominator zero, such value of the letter must be rejected and 
another must be selected in its place. 






13. 



is 



x-3 3x + 2 


23.T-9 


2-x 3: 


V 1 iC 1 


2 3 
7 v ff f 

< & JC, JC 


30 
17. 

19. 


5 ' 
5a-76 


2 10 
3 a 5 6 


8 12 4 


4 


6 


2a 4a Ga 


b + d 



20 . + _. 

xy xz yz 



01 be ,ac ab 

& \. ~j~ ' ^j " 

a b c 



ett . a b c a.b c 

**& : 1 h 



ab 



ac 



bo 



23. 



4a 



3a-7b 



5 be 3 jc 7 ab 
\ *. 

3 a 46 2c 



25 . 



6 a 



a 



26. 



m 



n 



m n m + n 



27. 



x-- 1 



2 + ^ 



X 2 -l 



60. 



[ 



3 a 2 ax 



a x a -f x or x 2 



29. 



a 



a b a- b 2 a -{-b 



30. 



32. 



a+6 a 6 



x2 xl 



81. 



4 x 5 



a a-- 



3x(a-b) 

-i L. c 

1 n to 



x 6 x-\-b x 2 b 



104 FRACTIONS 



+ 2) 2 

I I SUGGESTION. L. c. d. is ac(a +c). 

36. H - To divide 1. c. d. by denoins., see 

1-vO I O Cv "t " Ct/O -f~l -i / /"vr* 

Ex. 16, p. 93. 

37- ~^~~ ~~r H : r^ g TV 38. - -| - 



39. Write a rule for the addition (including subtraction) of frac- 
tions. (Apply 60 to each fraction before beginning.) 

(1) What is found first ? 

(2) What change is made in the several fractions ? How ? 

(3) What is done with the resulting fractions ? (See Exs. 1-8.) 

If precedes a fraction, what is done with its numerator 
(in parenthesis if it consists of more than one term)? If 
4- precedes a fraction, what sign precedes its numerator ? 

(4) What is the next operation ? 

(5) What reduction is very often made in the answer ? 

(6) How is the answer checked for accuracy ? 

11 21 x+6 
40 . 41 

3x-2y 5x-2y x-2 x + 2 or' + 4 



40 ^ I 43 _ 

^-t. [^ ^-*J 

x i. x x + a 



3 7 4 20 x SUGGESTION. Multiply both terms 

A^ . __ - . 

l 2x l + 2x 4 x 2 1 of the last fraction by - 1. ( 62, 1.) 



45. zz a_a-_-. > _ __. ( 62 

a_ 6 ^6 ft 2 -* 2 3_ a a 2 -9 



2 x 61 (i 9 

47 i . 48. h 

C 3 a; 4- 6 x 2 4-2x 3 a a 2 3 a 



ADDITION AND SUBTRACTION OF FRACTIONS 105 



49. - 4 ~ + p -^77-+^ * , 50. 

2-20 



SUGGESTION. As in arithmetic, in- 

i [ 9 ^2 

51. 3 # tegral quantities are given the denomi- 

x nator 1. 

SOLUTIO,. 



XX X X 

CHECK. Let x = 5. Then 3 x - 1 + 2a;2 = 44 j also = 44. 

x x 



52 . 3a! _ 



2a 



, 7 , - 

54. a + o 55. la 

a + b 5 - 6 a 

56. L-JL+a 1. 57. 



. 

2a a 1 

58. X 2_ a . + 1 -- 1 . 59 ^_ a _ 

a? + 1 a; + a 

60 . 4^-2-i^. 61. x*-3x- 3x ( 3 - 

2x+l x2 

62. a 2 -2aa;--^--f4a5 2 . 63. l 



64. l-[a-^_). 65. 



Moj,a;.2 
66. 5 # 1 \-x 2 . 67. H- 2 



^v ** ^ 

7 9 x-1 x+1 



68 . _ 3 + 69> _ 2> 

oj-f- 1 a? 8 - -1 (a + b) 2 a + b 



70. _ - _ H-- _ - 71. _ 4-- 

2 9 



106 FRACTIONS 



* 72 . ___ 73 m , y . 



1+m 



a; 2 - -2 a? 8 a^ + 5aj-4-6 y(xy) m(yx) my 



74. + +_ __ 75. 



b 2 ax ab a 1 a 3 

76. __ + =i_^- 77. 



3a 1-a 3 



xy 



4 a- - 9 b 2 9bd-6ad 



78. 



_ . - /^^A 

79 ' ^3 -- i --- ^3 -- i -- ( oO. 

or a 6 , or + a 6 



80. 



81. 2 ^ +^-i 



82 



83. 



84. 



85. 



, 

'-- 



+ 4 a; +2 1- 



4-7a-2a 2 4-5a-6a 2 

5a 2 a + 3 & 4a 



9 a 2 - 25 6 2 6 
86. ^- ^ 



SUGGESTION. Add 1st and 5th terms, and 2d and 4th terms, first. 



REDUCTION OF IMPROPER FRACTIONS 107 

88. ^ '- 1 

(a b~)(x a) (b a)(xb) 

b 4- c 
SUGGESTION. Change 2d fraction to See 62, 5, 3. 

(a - b)(x-b) 

89. , 2a +y .. + + 6 +v +*- - 



(x a) (a b) (x b) (b a) (x a) (x b) 
90. H 

\. Decomposition of a Fraction. It is sometimes desirable to 
reverse the operation of adding fractions. 

adn 4- ben bdm 
i. 

bdn 



SOLUTION " ~~ 4. 

bdn bdn bdn bdn 



= 1 + 3--' Ans ' ( 60 -) 
bdn 



6 a 2 3 b 2 + 10 c 2 a5c -f 6c^ + adc + abd 

z. - o. - - 

30 abc abed 



66. Reduction of Improper Fractions to Integral or Mixed Quan- 
tities. An improper fraction is one that can be reduced to an in- 
tegral or a mixed quantity ( 63). A proper fraction is one that 
cannot be so reduced. 

In exercises 51-69 of 64 we reduced mixed quantities to im- 
proper fractions by writing 1 for the denominator of the integral 
quantities and adding. In this article we reverse the operation 
of adding. 

Since a fraction is defined ( 59) as an expressed division, an im- 
proper fraction can be reduced to an integral or a mixed quantity by 
simply dividing its numerator by its denominator. 



108 



FRACTIONS 



1. 



a 2 +26 2 
a b 



2. 



SOLUTION. 



3 ^ _ 7 3. _ 



SOLUTION. 



a 2 - 


6 


a 5 


2 


a + 64- 3b< 


ab + 2 & 2 
a& 6 2 


c* ( c/ (- 

a 


6 





3a 
3a 


2 -7z- 10 

2 O v 

t t/ tJU 


x-3 


4 


3r | Q 
X -f- ^ 


-f 2a -10 
2x - 6 


x- 3 





362 _ 4 

The remainder is written over the divisor and the resulting fraction added 
to the quotient. (See 63, .) In the second example, after 4 is written 
over the denominator, the sign of the numerator is changed from to + and 
the sign before the fraction from -f to . ( 62, 2.) Algebraists regularly 
do this. 

In reducing fractions to mixed quantities, the rule is to continue the divi- 
sion as long as the divisor is contained an integral number of times in the 
remainder. The answers can be proved correct by addition. 



3. 
5. 
7. 
9. 
11. 



3 a; 2 + 2 x + 5 



x-3 

2x 2 -7x-l 
x-3 

a 4 -16 



a? + 2 
a 2 ax 3 x 2 

'' . m 4 

a-j- x 



1 



15. 



17. 



4. 



6. 



8. 



10. 



2a-3b 



2x-3 



5 a; 



+ 



bx 

/v.2 



a 



a 



14. 



16. 



18. 



10 a 2 -17 ax + 10 a; 2 
5 a x 



2x-3 



MULTIPLICATION AND DIVISION OF FRACTIONS 109 

67. Multiplication and Division of Fractions. 

, A AC M P MQ MQ 

Typeforms: X- = 



35 _ 2 v 7 25 

' 4 X f 2 '5 X 8 



4 S x - SOLDTION - v x= Ans - 

a 
CHECK. Let a = 4, 6 = 3, c = 5. 

2 a 6 6c 8 90 , 4 c n 



a cx 2 3 qfi 3ac 3a 2 5 14c<i 3 

X " X "" X ' 



8am 2 967i 3 l^xl^l/x?^ 3 10 7 a; Soft 8 

27 6V 16 am* 10/ 7z" 9x' c 3 3 y ' 



8 # 4 y _._ 2 a; 3 SUGGESTION. Invert the divisor and multiply as in 
15 aW " 3atf arithmetic. 



. ^_ 3x . 15 aa; 4 

' 5 ar ' 10 x 2 / ' ' ar 2 * ' 2* ' 26 

g 8 



SOLUTION. ~ x 



~ 1) (^ ~ 6)Ca>-q * - 7 ^ -f 



a: 2 



70 9 7 14 7 

CHECK. Let = 8. Then, x = - - ; also = - 

8 80 32 64 32 



a: 2 - 25 



a 2 5 a + 4 a 2 



xy 



ab-2b 2 3 -^ 

^* 



a 3 



110 FRACTIONS 

22. Make a rule for multiplication of fractions. 

(1) What operation is performed first, if necessary, with thu 

terms of each fraction ? 

(2) What is done with factors common to any numerator and 

any denominator ? 

(3) After canceling what is done ? 

(4) What is the last step if there is any uncertainty as to the 

correctness of the answer obtained ? 

23. On what principle does canceling factors common to any 
numerator and any denominator depend ? Notice that the prin- 
ciple applies only after the numerators are multiplied together 
for a new numerator and the denominators for a new denomi- 
nator. But it is more convenient to cancel before multiplying. 



25 

* 



ex (x -y)(x 



26 x - 27. xx 

a 6 & 6 a-* -- a 2 + 4 x 2' x3 X-- 



29 

x 



^ + 3a^ . a? + 3 a + 1 . a'-l. 

- - -- : -- - - - - * OX. -r- 

a; + 4. a^ + 4o; a a 

32. ?!nA 2 -5-(a-6). 33. ^ -*- (a + c). 

a; + y b + c 

34 8 a 3 . 4a 2 a 2 -121 a 

^ 



a 8 --4 

q? 

' 



8a; + 5 a 8 -! 
-7 x 2a;-l 



m 2 ?i + 2 m?i 2 + n 3 m + n 12 aj 3 -+ 24 x 2 x? + 2x 

a 4- X 4 t g* x + flJ rt g + 2o-15 . a 2 + 9a + 20 

a^"" a 8 -a? 8 ' cr + 8a-33 ""a 2 + 7 a--44' 



POWERS OF FRACTIONS 111 



42. Make a rule for dividing fractions. 
j3 + y/V/ . p 2 



r.s s 2 r -- s 4 ar 9 4 ar 1 

a-s _ 6 or 9 + 36 a? a; 4 + 21G a; 

-- 



45. 



: x 4 -17or 9 +16 a*~3x-4: 

^z D 



9a 4 -34 2 + 

3.2 _ 3. _ 12 x 2 2x 3 x 2 -\-x SUGGESTION. Invert 

47. X : -f- - divisor and multiply 

rpZ 11 nt- / /v> V y I '/ 

three factors together. 

a 2 ax a + x a 3 8 8 a 3 

48. - x - r-- 

2 3 a + a 2 a?- -a? a 2 + a# + ar 



49 -6 3 3 



2ab-2U 2 

a? 6xy + 9y 2 . (x* 9y 2 . a; 2 4- %y6y' 
y' 2 \x 2 + y~ x* - .'// - - 6 ?/ 2 



51 -. x x 

' ' / 



-xy + y x* + xy + y' 2 x y 



8*> HO -i / 9 / h 1 ' H/>9 

x- 2x 1 ox 1 6 ar/ ox L bar 

2 m 2 - -m 6 v 2 ??i 2 - m 3 2 m 2 - m -- 3 

5o. - X 



m 2 - 5 m + 6 6 m 2 - 11 w + 3 3 m~ - 10 m + 3 

2a 2 -5a-12 . 2a 2 -7a-4 
X ' * 



55. ^_ v^ w_^ 

68. Powers of Fractions. 

Type form : ( I = /T>* 

V bj 6** 

/ +j OLO C \ 4 4 Ct '.' _ / * wu \ rt 

1. I 1 Ans. r 2. f -3- - 3. 

' '' / 



112 FRACTIONS 



/ 2a*x\* (3 wm) 4 2 by 

4. I -- I 5. I -- I 6. * ^ 7. - 

V 7b 3 c 2 J \ 5byJ 7 bo (4c 3 ) 3 

(x 2 I) 3 SOLUTION. Notice that if (6) = a*b*, then 

8 ' (x-iy' (* 2 -l)3=(z+l)3(* 

Thus, Og^ = (*+l) 3 (*-l) 3 = (*+l)*. ^ 



. 



#12. O 4 ~ .V 4 ) 8 



15 x 16 

' 



69. Exercises involving Additions and Subtractions First, and Mul- 
tiplications and Divisions Afterwards. As a rule, before beginning 
the operation of multiplication or division of fractions, each factor 
should be reduced to a simple fraction, that is, to a fraction with 
integral quantities for its numerator and denominator. 



SOLUTION, c 2 + ^ - 2 c^ c-^. 
c 2 + d 2 c 

c d 

c tf-cd 



CHECK. Letc = 4, d = 3. Then 
j 2 a? \ . /., d\ 1.1 4 



c 2 + 1? 2 / \ c/ 25 4 25 c 2 + (Z- 25 



x \y~ y x 



X \ X\ f X 1 

* ^ : -- r 



1 + # a; y V 1 + x x 



ADDITIONS AND MULTIPLICATIONS 



113 



5. 



6 14 



25 x - 29 



2-ajJ (5-4a?)(2-a;) 



6- --8<M 

v a? 2 y 



-136 2 

' 



--*- 



a 



a? 



8 



a-3b 3b-a 3 a-3b 



r/ 1 
8. -f 



9. 



x x 



?/ 



. ^ i ~ 



X 



10. 3^-5- 



*ll. 6 



12. A- 



13. 



+ ay 1 a 2 



14. 



15. 



2 (x - 2) a-" 



y y \y 

L.2x + 4- 



a;-2 



3 a 4 



ax 3by 3bx + 



16. I?- 



x a + x a x\ [a x a + x 



17. 



a?b 2 a 2 + 6 2 a b 



114 FRACTIONS 

70. Complex Fractions. A complex fraction is a fraction that 
has a fraction in one or both of its terms. 

SUGGESTION. This means evidently that - is to be di- 
a b 

b vided by - Thus, a complex fraction merely furnishes a 

1. d 

c_ problem in division of fractions. Notice that the line of 

d the complex fraction must be made heavier or longer than 
the others, to avoid ambiguity. 

a 

SOLUTION. _5_ = +. = x * = 1 
c b d b c be 

d 
CHECK. Let a = 2, b = 3, c = 4, d = 2. Then, the given complex fraction 



ajg ^ . Q _ ] 


L . olo ^ ac?_2 x 2 _ 


i 




3 3 be 3 x 4 3 
3 cy 3m 2 2 #?/z 


a 


K r/ 2 r 

<J tl> a/ 


4?^ 


5 a 2 6 


b 


9 




y| 


5. 


26 


. o 

ora 


3 x 2 y 


. 

c 


7 cm/ 


7n 2 


4a% 




+ & 


^- ? / 2 


a 3 - 6 3 


i+i 








a b 


x 


a + & 


a; 


6. 


*t 




o 


, * 
a-- b 


X + ?/ 


a 2 6 2 


1 

1+ 


_ 






a + o 


y 


a 


2/ 



SOLUTION to Ex. 9. *I - ^+_1 = I x __!_ = ^_+_y . ^ ng> 

x y a: y + 1 xy + x 





a- ^ 


1 


+ S 


m 


X 




10. 


C 


11. 


6 


n 


y 










a + - 


1 


+^ 


- + 


m 






c 




a 


?i 


y 






a a + 6 


i 


-3y 


a; - 


2 x 


+ 3 


13. 


6 a -- b 




i + ?/ 


X 
1 ^ 


3-> 
JU 


+ 4 


b a -- b 




1 


-I- 


"* 



a a + 6 ;--2/ a;+La; 



COMMON ERRORS IN FRACTIONS 115 

6 a-\-b b a; -r 1 a?-- 1 

/y _ 1 I _ J ^1^ __ L 1 

x 6 b a + b , as - - 1 a? -f 1 

*lo. ! 3 

S-2 + ^7 +, 

x o a o 

1 
2 m 3 -f 

wi 




971 X -^ Cl 

1 REMARK. Such a fraction as this is often called a 
22 _ _ _ . 

i continued fraction. To simplify it we commence at 



~T the bottom and replace each complex fraction by its 

- equivalent simple fraction. 
x 

SOLUTION. - - - = - - - = - ^ - = x + l. Ans. 

x + l-x 




x 

a. Common Errors in Fractions : Certain errors are made so often and 
by so many pupils that it is desirable to describe them, and caution against 
them. Checking serves to find such errors. 

1. Never cancel single terms in the numerator and the denominator of a 
fraction when either is a polynomial. Only factors of the whole numerator 
and the whole denominator can be canceled ( 00). 

2. In subtraction of fractions, and later in equations, when a fraction is 
preceded by , students must change the signs of all the terms of the nu- 
merator in simplifying. 

Thus, q_ zL d = a-c + d 

b b b 

?>. To change the signs of both numerator and denominator, the signs of 
all the terms of each must be changed. 

Thu 3& - a _a 36 (This amounts to multiplying both terms 

'2d^~<r^2? b y -i.by6o.) 

4. To divide both terms of a fraction by a monomial, every term of both 
numerator and denominator must be divided by it. 

5. In addition of fractions, after they are reduced to a common denominator 
and the numerators are added, the sum must be placed over the common 
denominator. In equations, denominators disappear (see 85). 



116 FRACTIONS 

71. Miscellaneous Exercise in Simplifying Fractions. Check 
answers and be careful to see that answers are in their lowest 
terms. 



3 - 



a 



l x-l x-3 

4. - ! - 5. - -- 6. 



oj~2 a;--l a--l a 3 -! 



g ab z -b 3 a?-ab 
2 x 



a - # a + a 2 6 2 or - y 2 

a 3 a 5 4 a % 3 a; 2 + 2 a; + 1 

~2 6 8~' ^Tl X a^ 



12. _= x _. is. + i + 

or-}- ay \a? 

/I - a\* a 2 + 6 2 

1 ( -) a 6 - 1 

14. _ V 1 + ^ . 15. _ 1 



a + 



2 # 2 + 2 cc - - 15 x + 5 x 2 - - a 2 x a 

l-2x l-2x-x 2 + 2x 3 1 3mn 



3 



1 -}- X s 1 -{- 3ic -f- 3# 2 + it* 3 n-- m n* m 

20. -^ + -1 ___ t_. 21. -5 ___ _ 11 

/> /^'/> /i <>9 o O7ir ji 



6a + 6 6-6a 3a 2 -3 26+2 46-4 6-66 3 



22. x- + /--... 23. 



+ ?/ a 4 if 



1 \ / r 

24. (a 4 - :: ]-i-[0-V 25. 

nA I \ nt 

Jj \ A/, 



RATIO AND PROPORTION 117 

72. Ratio and Proportion. 

1. The ratio of two numbers is the quotient of the first number 
divided by the second. The ratio of 3 to 8 is f . We see from 
this that a ratio is & fraction. 

2. What is the ratio of 3 to 4? of 6 to 8? of $7 to $14? of 
20 to 10 ? of 9 men to 3 men ? of 20 Ib. to 5 Ib. ? of 12 ft. to 1 ft. ? 
of i pt. to 1 qt. ? of I to | ? of a dollars to b dollars ? of c to 
w + n? of cr-6 2 to a-b? of m 2 -9 to m 2 + 6 m + 9? 

3. The first number in a ratio is called the antecedent, and the 
second the consequent; both are called the terms. A ratio is an 
abstract number, since it is always the quotient of one number 
divided by another of the same kind. A colon is often used to 
denote a ratio. Thus, 4 : 7 signifies the ratio of 4 to 7. 

4. A proportion is an equation ( 42) the members of which are 
ratios. 

Thus, f = f j or, 3 : 4 = 6 : 8 is a proportion. 

5. Fundamental Tlieorem. In a proportion the product of the 
"extremes" or first and last terms, equals the product of the " means." 

PROOF : If - = - is any proportion, 

fiw a ^ HA c M (Because, equals multiplied by bd give 

men, x />a x oyc, 

p ft equals.) 

(a and d being the extremes, and b and c 
or, ad = be 

the means or middle terms.) 

6. Solve for x in = , or 16 : x = 12: 27, using theorem. 

x 27 

Q i P 9 6 45 x x $12 

7. Solve for x :-= ; = --: = ; 

x 24' 60 24' 21 $36' 

12 Ib. : 20 Ib. = 15 cents : x cents ; $ 6 : $ 15 = 75 yd. : x yd. 
(a) As ratios are abstract, drop $, ^, etc. , before solving. 

8. If 12 T. of hay cost $ 132, how much will 16 T. cost ? 
SOLUTION : x = f x $ 132 = $ 176. (By reducing 16 : 12 to 4 : 3.) 
Notice that the answer x will always equal the number correspond- 
ing to the answer multiplied by the proper ratio of the other two. 




118 FRACTIONS 

9. If 24 bbl. of flour are eaten by a garrison in 9 weeks, how 

many barrels will be eaten in 4 weeks ? 

w 

iHiswBwBi ' p 10. A lever is a bar having two 
^ forces, the power P and weight W, ap- 

plied to it, which counterbalance each 
other. The point of support, F, is 
called the fulcrum. " 

11. It is always true of a lever that the ratio of the power to the 
weight equals the ratio of the weight's distance from the fulcrum to 
power's distance from the fulcrum, or P : W= b : a. 

12. Find P in a crowbar when W= 90 lb., a = 4 ft., b = 1 ft. 

13. Find Win an oar when P = 25 lb., a = 10 ft., 6 = 6 ft. 
Fulcrum is where blade is in water, and TF"is at rowlock. 

14. Two horses pull on a lever 44 inches long with forces equal 
to 300 lb. and 350 lb. Where shall fulcrum pin be put so that the 
forces balance ? Let x and 44 x equal lever arms. 

15. Two weights of 12 lb. and 8 lb. respectively at the ends of 
a bar 14 ft. long (supposed without weight) balance each other. 
How far is the fulcrum from the 8 lb. weight ? Let x = number. 

16. One quantity is said to vary as another when as one 
changes, the other changes in the same ratio. For example, 
the total pay a man gets varies as the number of days he works. 
If he works twice as long one time as another, he gets twice as 
much money. If two quantities vary so that the ratios of old to 
new values are the same, they are said to be in proportion. Thus, 
if 5 bu. of wheat are worth $5.75, 8 bu. are worth $9.20, since 
5 bu. : 8 bu. = $ 5.75 : $9.20. 

The word " vary " is also often used in a loose way to denote that 
as one quantity changes another quantity dependent on it also 
changes, though not in the same proportion. Thus the prices of 
articles vary with the seasons, etc. 



GRAPHICAL RELATIONS 



119 



* 73. Graphical Relations. If two quantities are related, the 
change in one as the other changes can be pictured to the eye by 
means of " squared paper." Squared paper is paper accurately 
divided into little squares. (See page 172.) 

1. To show graphically the price of butter throughout a year. 

SOLUTION. "We take a piece of squared paper and write Jan. 1 at its 
lower left-hand corner (see figure below) ; Mar. 1 at the right of Jan. 1 and 
below the second heavy vertical line ; May 1 below the third heavy vertical 
line ; and so on to the following Jan. 1. (Two-month intervals are taken 
instead of one-month intervals to save space.) We now count upward from 
the lower left-hand corner 83 spaces (for 33^, the price of butter on Jan. 1 
see table of prices given at the left of the diagram below), and put a point 
at a. On the second heavy vertical line we count upward 34 spaces (for 34 ^, 
the price on Mar. 1), and put a point at b; and so on for each month. 
Through these points we now draw a running line, abcdefg. This line gives 
us very quickly an idea of the changes in the price through the year. If we 
suppose the changes in the price are made gradually, we can find the prices 
at any intermediate dates. Thus, on June 1 the price was 2G/-. 



Wholesale Price of Butter 

In New York City, 
1907-1908 














b 






















































-^- 


- 






\ 




















































( 










\ 






























































\ 
















































3( 


(' 












\ 






























































\ 










































</ 


Month 


Price 






















X 






























































\ 






































/ 
























x 








/ 








. - 




p 


~~. 














/ 




Jan. 1 
Mar. 1 
May 1 
July 1 
Sept. 1 
Nov. 1 
Jan. 1 


330 
3i0 
28^ 
250 
26 
240 
29^ 


a 
b 
c 
d 
e 
f 



25 C 
























\ 




^v 












/ 


















































^ 


N 






^ 






















































/' 


"*~~ 


































































































































a 


(.' 










































































































































































































































































































15(2 









































































































































































































































































































2. Make a table 
like that for the pre- 
ceding problem from 
the following whole- 
sale prices of eggs in 
K Y. City in 1907, J 


10^ 










































































































































































































































































































.-> 


- 






































































































































































































































































































































































n. 1 Mar.l May 1 Julyl Sept.l Kov.l Jan. 1. 



120 



GRAPHICAL RELATIONS 



and then construct from the table a diagram, using graph paper: 
Jan. 1, 36^; Mar. 1, 23^; May 1, 21^; July 1, 21^; Sept. 1, 
30 ^ 5 Nov. 1, 45 ^ ; Jan 1, 34 



3. Make a table and the corresponding diagram from the fol- 
lowing prices of wheat in 1907. Jan. 1, 73 / ; Feb. 1, 79 ^ ; Mar. 
1, 79^; April 1, 81^; May 1, 96^; June 1, 97^; July 1, 94^; 
Aug. 1,93^5 Sept. 1, 100^; Oct. 1, 104^; Nov. 1, 93^; Dec. 1,96^. 

4. Make a table and diagram from the following Chicago prices 
of oats (1907-1908): Jan. 1, 34^; Feb. 1, 37^; Mar. 1, 40^; 
April 1, 42/; May 1, 45^; June 1, 49^; July 1, 42^; Aug. 1, 

Sept. 1, 54^; Oct. 1, 52^; Nov. 1, 49^; Dec, 1, 

Jan. 1, 



<f 



5. Make a 

of prices from the 
accompanying dia- 
gram showing the 
minimum price per 
cwt. of sheep in 
Chicago, 1907. 

Notice that one 
side of small square 
represents 20^. 

6. The following 
data give the prices 
throughout a cer- 
tain year of the stock of the Atchison, Topeka, & Santa Fe railroad. 
(The 80 means that one $100 share is worth $80 in money on 
Jan. 1.) Make a table and diagram from these data: Jan. 1, 80; 
Feb. 1, 75; Mar. 1, 76; April 1, 79; May 1, 83; June 1, 81; July 
1, 84; Aug. 1, 92; Sept. 1, 9G; Oct. 1, 91; Nov. 1, 87; Dec. 1, 83; 
Jan. 1, 85. 

7. Put the diagrams of Ex. 8 and 9 together on one sheet, and 
see which increased more rapidly, wages or the price of beef. 



Jan. 1 Mar.l May 1 July 1 Sept.l Nov.l Jau. 



GRAPHICAL RELATIONS 



121 



8. The following data give the average wages per hour in 50 
selected trades from 1898 to 1907. Make a table from them and 
then a diagram. '98,22.5**; '99,22.9**; '00,23.6**; '01,24.2**; 
'02, 25.1**; '03, 26.1**; '04, 26.3**; '05, 26.7**; '06, 27.9**; '07, 29^. 

9. The following data give the retail prices of sirloin steak in 
New York City from 1898 to 1907. Make a table, and then a dia- 
gram from them. '98, 18.3**; '99, 18.4**; '00, 18.7**; '01, 19.4^; 
'02, 20.4**; '03, 20.3**; '04, 20.7**; '05, 20.8**; '06, 21.3**; '07, 22**. 

10. *The average rainfall in inches in Chicago for 32 years was 
as follows: Jan., 2.1; Feb., 2.3; Mar., 2.6; April, 2.7; May, 3.6; 
June, 3.8; July, 3.6; Aug., 2.8; Sept., 2.9; Oct., 2.6; Nov., 2.7; 
Dec., 2.7. Make a diagram, using centimeter as unit. 

11. The length of the day in hours on the 21st of each month 
at Washington, D.C., in 1908 was : Jan., 9.9 ; Feb., 11 ; Mar., 12.2 ; 
April, 13.5 ; May, 14.5 ; June, 14.9 ; July, 14.5 ; Aug., 13.5 ; Sept., 
12.2; Oct., 10.9 ; Nov., 9.9 ; Dec., 9.4. Make a diagram from data. 

12. The course of a ship 
in the Gulf of Guinea ran 
through the following points 
located by their longitudes 
and latitudes : Mon. noon, 
2E., 2K; Tues. noon,lE., 
1N.; Wed. noon, 1E.,0N.; 
Thurs. noon, E., 1 S.; Fri. 
noon, 1 W., 1 S. 

13. The log of a ship read 
as follows : Mon. noon, 4 E., 

3 N.; Tues. noon, 3 E., 1N. ; Wed. noon, 1E., 0N. ; Thurs. 
noon, 1 W., 2 S. ; Fri. noon, 3 W., 3 S. Make a diagram show- 
ing the course of the ship. 

14. Squared paper can be used also to make statistical dia- 
grams of heavy lines, as found in geographies. Make several. 

*For similar data use New York World or Chicago Daily News Almanacs. 



M<m, 













1 d W 











F 








1 


K. 






/ 


i 


Ej 










































/ 
















































/ 














































> 


/ 
















































/ 
















1 


X. 
































T 


it- 


-. 




















































































































































































































n 


N. 






























/ 


\\ 


V 


1. 












































/ 
















































/ 
















































/ 














































^ 


X 
























1 


S. 








K 


n 












T 


111 


IS 



































































































































































































































CHAPTER IV 
SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

74. An equation in algebra is a statement that two quanti 
ties have the same numerical value ( 24). 

75. Equations are of two general kinds, identical and conditional. 

76. An identical equation either contains no letters, or its let 
ters may have any values. The sign for an identical equation 
is =. 

Thus, in (x + 2) (a? + 3) = x 2 + 5 x + 6, 

show that the equation holds true for x = l, x = 2, x = 5, x = 0. 
etc. All given quantities in what is called literal arithmetic set- 
equal to their answers make identical equations. By literal aritli 
metic is meant addition, subtraction, multiplication, division,, 
powers, roots, factoring, and fractions in which letters as well an 
figures are used to represent numbers. The student will recall 
that we have been checking our answers in literal arithmetic by 
assigning any values to the letters. 

Algebra is often defined as the science of the equation. By 
this definition literal arithmetic belongs rather to arithmetic. 

REMARK. The sign = is commonly used for identical equations as well 
as for conditional equations ( 77). Thus, so far, we have always used =.. 
Whenever it is desired to draw special attention to the fact that an equation 
is of the identical kind, or that one letter is to be used to take the place, o^i 
some combination of other letters to save space, the sign = will be used. 

77. A conditional equation is one that is true only for cer* 
tain values of the letters involved. Conditional equations appeav 

in the solution of problems. 

122 



DEFINITIONS 123 

The equations obtained from the problems of 1, 31, and 42 
were conditional equations. Ordinarily when we speak of equa- 
tions we mean conditional equations. 

78. A simple equation is one which, when reduced to its simplest 
form, contains only the first power of the unknown quantity or 
quantities ( 57). 

79. Known and Unknown Quantities. The custom in algebra is 
to represent unknowns, as a rule, by the last letters of the alpha- 
bet, and known quantities by the others. In physics and other 
sciences the custom is to denote quantities known and unknown 
by initial letters, as v for velocity, p for pressure, t for time, etc. 

80. The root of an equation is the value of the unknown quan- 
tity, that is, the answer. This use of the word should be dis- 
tinguished from root of a quantity ( 18). The student may try 
to explain why the answer is called the root of the equation. 

81. To solve an equation is to find its root. The word " solve ' : 
means literally to "loosen' or "untie." When one solves an 
equation, he loosens or unties the unknown quantity x from the 
others, and gets it by itself on the left side, or " member," of the 
equation. 

82. The verification of an equation is accomplished by substi- 
tuting the root found in the original given equation and getting an 
identity. If a value of the unknown quantity, upon being sub- 
stituted in the given equation, makes the two sides reduce to 
the same number, after simplying, this value is said to satisfy the 
equation. 

83. An axiom is a truth assumed to hold in a study or in- 
vestigation. The axioms have been called self-evident truths 
because they are so simple that the mind accepts them as true 
directly. The use of axioms changes the study of equations into 
a science. 



124 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 



AXIOMS 

1. ADDITION AXIOM : The same quantity, or equal quantities, 
added to equal quantities, gives equal quantities. 

(1) 4 + 8 = 12 (2) 5 + 11 = 10 + 6 

2 = 2 3+ 5= 7 + 1 



4 + 8 + 2 = 12 + 2 5+11 + 3 + 5 = 10 + 6+7 + 1 

2. SUBTRACTION AXIOM : The same quantity, or equals, sub- 
tracted from equals, leaves equals. Thus, 

(1) 3 + 9=5+7 (2) 6 + 11 = 2 + 15 

6=6 2+ 3 = 4 + 1 



3 + 9_6 = 5+7-6 6 + 11-2-3=2 + 15-4-1 

3. MULTIPLICATION AXIOM: Equals multiplied by the same 
quantity, or equals, give equals. 

15 + 10 = 30 - 5 
x2= x2 



30 + 20 = 60 - 10 

4. DIVISION AXIOM : Equals divided by the same quantity, or 
equals (not zero), give equals. 

5. POWER AXIOM: Equals raised to the same power give , 
equals. 

6. ROOT AXIOM : Corresponding roots of equals are equal. 

7. Quantities equal to the same quantity are equal to each other. 

8. GENERAL AXIOM : If the same operation is performed on two 
equal quantities, the results are equal. 

The last axiom holds true in general only for arithmetical 
numbers. 

The axioms will be referred to as Add. Ax., Sub. Ax., Mult. 
Ax., etc. 



SOLUTION BY MEANS OF AXIOMS 125 

84. Signs for Words. As solving equations is reasoning, we 
shall need signs denoting cause and conclusion. Mathematicians 
use .'. for "therefore" or "hence," and v for "since" or "because." 

85. Solution of Equations by Means of Axioms. Very simple 
equations can be solved by ordinary reasoning without using the 
axioms; but when the equations become at all complicated, it is 
much easier to make use of the axioms. 

1. Solve - + ^ ? = - 3 and verify. 
326 

SOLUTION. 1. c. d. = 6. 

.-. 2 x + 3 x - 6 = 7 x - 18. (Mult. Ax. The equals, f + 2Ln2 and 

7 x 32 

3, multiplied by the same quantity, 

6 

6 (the 1. c. d.), give the equals 2 x -f 3 x 6 
and 7 x 18). 

.*. 2x + 3x 1x = 18 + 6. (Sub. Ax. (see p. 61). --The same quan- 

tity, 7 x 6, subtracted from the equals, 
2 x + 3 x 6 and 7 x 18, gives the equals, 
2x + 3xlx and 18 + 6). 
.-. _ 2 x = - 12. (By addition, 29, Ex. 24-32.) 

.*. x = 6. (Div. Ax. The equals, 2 x and 12, 

divided by the same quantity, 2, give the 
equals, x and 6.) 

VERIFICATION. + <L=J* = !^>_ 3; or, 2 + 2 = 7-3. 

Q A O 

REMARKS, (a) The first operation, in which the equation is multiplied 
through by the I.e. d., is called clearing of fractions. Notice that in the 
new equation the denominators have disappeared. 

(&) The second operation, in which all the unknowns are placed on the 
left side of the equation and the known numbers on the right side, is called 
transposition. (See page 61.) To transpose any term, that term with its 
sign prefixed is subtracted from both members of the equation. Any term 
can also be transposed by adding the term with its sign changed to both 
members. 

(c) Simplifying members by adding is often called '"collecting.''' 1 

(d) The last operation is called dividing through by the coefficient of x. 



126 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

Because it is a little difficult to describe the use of the axioms 
on the preceding page, we will explain this matter more in detail. 
It is very important that the student should understand this sub- 
ject, as it is fundamental in algebra, and his ability to solve 
equations depends in no small degree on his knowledge of the 
axioms, and of how they are applied. 

(1) Explanation of Clearing of Fractions. 

- + ^^ = I* - 3 (equals) 
326 

multiplied by x 6 = x 6 (the same quantity) 
give 2x + 3x 6 = 1 x-- 18 (equals). (Multiplication Ax.) 

(2) Explanation of Transposition. 

2x + 3x -6 =7318 (equals) 

diminished by _ 7 x 6 =7 x 6 (equals) 

give 2x + Sx-~7x -18+0 (equals). (Subtraction Ax.) 

(3) Explanation of Dividing through by Coefficient of x. 

- 2 x = 12 (equals) 
divided by -+--- 2 = -4--- 2 (the same quantity) 

give x = 6 (equals). (Division Ax.) 

Solve and verify, and explain the operations, especially those 
based on the axioms in the following. Write all the reasons at 
the right margin, as is done in the model. In oral explanations 
the student should point to the appropriate members of the differ- 
ent equations as he repeats the axioms. 

x 2 x -f- 3 x 2 x + 7 x 7 

= - H -- o. -- -- = 

2343 341 

4-l 3_ 3 + 5 4x 5 3 8xQ 

~ ~~ ~' 5> ~~ = ~~' 



6 . <_: = 10 + =- 7. . 



AXIOMATIC CHANGES 



127 



8. 



_ A 
10. 



x + 2 x 3_ 
3 4 



.2-*- 1 



2 



9. 

a; 





= 2 a; 



11. 



65 



3 a; + 2 



-13. 



86. Explanation of the Axiomatic Changes made in Equations by 
Means of the Equation Balance. It is very desirable that the stu- 
dent should get the idea that an equation is very different from a 
quantity, though made up of two quantities. An equation is very 
much like two weights balanced on scales. In the one case the 
numbers on one side added give the same sum as the numbers 
added on the other side ; in the other case the weights on one 
side added give the same total weight as the weights added to- 
gether on the other side. 




Axiomatic changes in an equation can be easily and simply 
illustrated by means of a special kind of balance, called an equa- 
tion balance, one form of which is shown herewith. F is the base, 
a thick board ; ON is a vertical post ; PQ is a cross arm fixed 
rigidly at right angles to ON. BC is the scale beam moving 
about pivot at A. At P and Q are frictionless pulleys fastened 
at equal distances from N. The letters h, m, n, k, mark scale 
pans. For weights whole pieces of crayon can be used. 

Let us call the scale pan m on the 1 jft side positive ; then h 
is negative, since it pulls in the opposite direction. On the 
right side n is positive, and k negative. 



128 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

1. To explain transposition. Place weights on the pans so that 
the apparatus balances, as 4 on h, 8 on m, 6 on n, 2 on k. Thus, 



If now we remove 2 on & from the right side to the left side, we 
must place it on m if we wish the apparatus to continue balanced. 
Thus, 2 on the right side must become + 2 on the left side. In 
other words, when we transpose we must change the sign. 

2. To explain clearing of fractions. Place weights on the scale 
pans so that the apparatus balances, as 3 on h, 5 on m, 4 on n, and 
2 on k. Now double the number on each scale pan. Thus, 

-3+5=4-2 
becomes 6 -f 10 = 8 - - 4. 

The student should learn from this that in clearing of fractions 
he must be careful to multiply each term of the given equation by 
the lowest common denominator. 

The other axioms can be explained in a similar manner. 

Interest will be added if this apparatus is actually made and 
used. It can be employed to explain - times - gives + by de- 
scribing putting on weights as positive, and taking off as negative. 
Thus, if 4 crayons are on the h pan, and two are taken off twice, 
to balance the scales 4 must be put on the n pan on the other side. 
Hence, -2x-2 = + 4. 

87. Exercise in Solving Equations containing One Unknown Quan- 
tity. Solve the following, writing the reasons at the right margin 
as is done in the model, 85. Be sure also to verify the answer 
in the given equation. 

1. 9 x - 2 = 7 x + 16. 

SUGGESTION. As this equation does not contain fractions, the first step in 
the solution is to transpose (see 86, b). 

2. 6x-25=7 -2x. 

3. 8a-7 + 3a 



SOLVING EQUATIONS CONTAINING ONE UNKNOWN 129 

4. 6a; + 4a;-13-2a;-3 = 0. 

5. 42 -21 a; + 15 = 57 + a; +22. 
6. 



SUGGESTION. Perform the multiplications, and then transpose the un- 
known terms to the left side, and the known terms to the right side. The x* 
terms cancel. The equation is now like those we have solved. 

7. 3(x - I) 2 - 3(x 2 - 1) = x - 15. 
8. 

9. 2(a; + 2)(a;--4)=a;(2aj 

10. 9(13 - a?) -13 a? = 5(21- 2 a?). 

- ^ Ju i JU Ju **j ~\~ 10 ~~ -^ ~~~~ *-^ "~"~ ^ 

U ' 2 + 3~4 = ~2~' 12 ' "2 4~ = ^' 

( 64,39, (3): 85,1.) 

a 2 a; 3 a^-^7 x_ x 2_ x x 9 

~5~ 4 " 10 " 6 5 "5 4 

3r4- 1 *> 7- r 1 1711 

15. + f^ = 10 + - i. 16. ; + i=- 

.w O O O *v O *v 



17. ^j^ = 3 . 18. z^ + 24 = 

I 1 + 1 o 

3y 180-5y_ 29 2Q 60 -a; 3.T-5_3a; 
19 ' T + 6~~~ y * ^4" 7 " 4 

21. = a; ~ 5 f ,- 22. E 1 + 1 = 1. 

2a;~5 2^-2 ar + 1 x 

v q/v, ^4.i'^--Sr1--iB 

23. = ^L_2. 24. 2- P = ^ 

it' + l a; + 2 6 12 1 + a; 

^ / _l_ Q 9/y ^ 3ft 4^ 

25. ^i2 + ^Zl3_ 6 . 2 6. -^ -= - 

O^ -^ 2 O* - 7/ ^ ?/ 



130 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

27. Make a rule for solving simple equations in one unknown 
quantity answering the following questions : 

(1) If the given problem calls for multiplications (excepting those 

in denominators), what is done first ? 

(2) If the equation contains fractions, what operation is per- 

formed ? Upon what axiom does this change in the 
equation depend? 

(3) After removing parentheses, in case any remain, what changes 

are made in the terms of the equation ? Upon what 
axiom does this change in the equation depend ? 

(4) After transposing, what usually has to be done with the 

terms ? Does this depend on an axiom ? 

(5) What is done last of all to complete the solution ? Upon 

what axiom does this step depend ? 

(6) How is the answer verified ? 

28. *1_* = 3 = . 29. 



x i x + 3 x a? + 1 x 1 



3x-l 9x-31 321 

oO. - --- - - - . ol. 



874 z + 1 x+2 x+3 



32. - + = -. 33. 



6o;+4 2x+3 2x--l 3x- 6 

34. 



3x+5 6 2x+3 & \ 9 



36. _l_ = 7a , 37. - 



9 3 6|-3z S(x-2) 



38. _ = _ 2 _ 

3 w 2 



40. 3.2^-^^^55=8.9. 41. . 



.. . 

.5 .5cc .4 2 a; .1 



42. S-a-- 
x) 2 



43. 



SPECIAL PROCESSES OF SOLUTION 131 

8 2,-e-l 



2x-l 4ar -1 1+2* 

44. = 2 

v j J_ X J_ X -L 

45. ~~ fl? + rrL: 



a; 



46. 3 + = 

4 2 3/6 3 



47. t_l_*_4 = t- 



*48 - I ^ = ^ I 

j r-v 1 *~* /^ \ I 



49. 



50. 



4-2o; 8(1-*) 2-# 2-2o? 

6 



:(a?4 

Y" 2 x 2 1+3 a? 2 x 2- 



a - 



15 14 (#-1) 21 6 105 

51. (a? 5)(jB-2)-(o? 6)(2 6) + (a? + 7)(a? 2) = 0. 



Special Processes of Solution. The regular method of solu- 
tion is : (1) clear of fractions ; (2) transpose ; (3) collect ; (4) 
divide through by the coefficient of x. But any method is per- 
missible which solves, or loosens, x from the other quantities, 
provided the several steps can be justified. 

89 Problems and Exercises. By a problem in algebra is com- 
monly meant a question that involves first the forming of an equa- 
tion and afterwards its solution. See 1, 31, and 42. Other 
questions are usually called exercises, though any question calling 
for a solution or simplification can be called either a problem or 
an exercise. The word " exercise J; is also used as a collective 
noun, to denote a series of questions. 



132 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

90. Steps in the Solution of Problems. It is important that the 
student recognize clearly the several steps in the solution of prob- 
lems like those in 1 and 31. These steps are : 

1. Looking for the unknown number and letting x equal it. 

2. Constructing the functions of x, that is, the quantities con- 
taining x described in the problem. 

3. Turning a sentence in the given problem into an equation. 

4. Solving the equation by the method of 87. 

In the following articles training is given in constructing func- 
tions and in changing sentences into algebraic language. 

91. Exercise in constructing Functions. To test answers, sub- 
stitute for x any value, and work as an arithmetical problem. See 
whether the answer agrees with the result obtained by substi- 
tuting the same value of x in the algebraic answer obtained. 

Express the answers to the following in algebraic language : 

1. What is the value in cents of x dollars ? Ans. 100 it' ^. 

CHECK. Let x = 5. Then the question reads what is the value in cents of 
five dollars ? Ans. 500 ^. Also putting x = 5 in the answer above, we get 
100 x 5 = 500, or 500 f . 

2. What is the value in cents of x two-dollar bills? Of x 
quarters ? Of a? dimes ? Of x nickels ? 

3. If a man sold a horse for 3 a; -f 6 dollars and lost x 1 dol- 
lars, how much did the horse cost ? 

4. If a house is worth 5 x + 10 dollars and the lot it stands on 
3 x 1 dollars less, how much are both worth ? 

5. What is the distance around a rectangular field whose 
width is x rods and length x + 10 rods ? 

6. What part of 9 is 5 ? What part of 9 is x ? 2 is what 
part of x ? 

7. A room is x ft. long and x 5 ft. wide. How many square 
yards are there in its area ? 

8. By what number must 8 be multiplied to produce Sx 5 ? 



FUNCTIONS 133 

9. The smallest of four consecutive numbers, that is numbers 
that follow one after the other, as 7, 8, 9, 10, is x. What are the 
others ? 

10. Write down five consecutive numbers of which the middle 
one is x. 

11. The sum of two numbers is 15 and one is x 1. What is 
the other ? 

12. What is 15 per cent of x + 5 ? 25% of 3 x 7? 

13. The number x + 5 is 20 % of what number ? 

14. Express in cents x dollars and x -f- 5 cents. 

15. A boy had x dollars and spent x + 1 dimes and x -f 2 cents. 
How many cents had he left? 

16. If the length of the day is x + 4 hr., what is the length 
of the night ? 

17. A boy is #-f-4 yr. old to-day. How old was he 6 yr. ago ? 

18. A man is 40 yr. of age. How old will he be in x + 2 yr. ? 

19. The divisor is 4, the quotient is # + 1, and the remainder 
x: what is the dividend ? 

20. "Rate, time, and distance " problems occur very frequently 
in algebra, and the student ought to be skillful in constructing the 
functions needed in solving them. By rate is meant the number 
of miles per hour, or meters per second, or the like ; by time is 
meant the number of hours or minutes or seconds ; by distance, 
the number of miles or meters, etc., in the whole distance. It is 
possible to generalize from a single problem. 

If a man rides for 5 hr. in a carriage that moves at the rate of 
6 mi. per hour, how far does he go ? Keeping this problem in 
mind, answer the following questions : 

(1) When time and rate are given, how is the distance always 
found ? Ans. Multiply the time and rate together to get 
the distance. 



134 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

(2) When the distance and time are given, how is the rate 

found ? 

(3) When the distance and rate are given, how is the time 

found ? 

21. If a man travels x mi. per hour, how far does he go in 
10 hr. ? 

22. If a carriage travels x mi. going 4 mi. an hour, how long 
is it on the road ? 

23. If a train travels 180 mi. in x 4 hr., how many miles per 
hour does it go ? 

24. If a horse ran 2ic+3 minutes at the rate of 40 rods per 
minute, how many miles did it go ? 

25. If 60 3x is the distance and 2xl is the time, what is 
the rate ? 

26. If J x + 18 mi. is the distance one train goes and 33 mi. 
is its rate, and ^ x 18 mi. is the distance and 21 mi. is the rate 
of another train, what is the number of hours each is on the road ? 

27. If A travels at the rate of 7 mi. in 5 hr. and goes x mi., 
how long does it take him ? 

28. If x is the distance to the top of a mountain, and a person 
walks up at the rate of 1^ mi. per hour, and down at the rate of 
4i mi. per hour, how long is he away from home ? 

29. A man travels x mi. an hour for a hr., and then 2x 3 
mi. an hour for b hr. How far does he travel in all ? 

30. What is the velocity in feet, per second, of a train that 
travels 90 mi. in x hr. ? Of a train that travels x mi. in 
90 min. ? 

31. How many miles can a man walk in 30 min. if he walks 
1 mi. in x min. ? How many feet can he walk in a minute if he 
walks x mi. in 1 hr. ? 



FUNCTIONS 135 

32. " Number of articles, price, and cost " problems are often 
given. By price is meant the cost of a single article ; by cost the 
value of all. Thus the price of sugar is 6 ^ a pound, while the 
cost of 11 Ib. is 66^. 

(1) When the number of articles and price are given, how is the 

cost found ? 

(2) When the cost and number of articles are given, how is the 

price found ? 

(3) When the cost and price are given, how is the number of 

articles found ? 

33. A man bought x tons of hay at $12 a ton. How much did 
he pay ? 

34. Fifty yards of cloth cost x dollars. What is the price per 
yard ? 

35. Each of x 2 persons paid $4 for a trip. How much did 
they all pay ? 

36. A man paid $2 a day for board for 26 x days. How much 
did his board cost him ? 

37. If 2 x 5 Ib. of tea cost $9, what is the price per pound ? 

38. What is the total cost of x dozen oranges at 40 ^ a dozen, 



and 25 x dozen at 35 a dozen ? 

39. If S.E was the cost of a camping trip, how much more 
would each man have to pay if there were only 8 men instead 
of 10? 

40. If x eggs cost $ 1.05, how many cents will 7 cost ? How 
much more will 9 cost than 7 ? 

41. If maple sirup sells at $1.40 a gallon one month and 
$ 1.60 another month, how many more gallons could you get for 
$# the first month than the last month? 

42. Length, breadth, and area problems occur frequently. 

(1) How is the area of a rectangle found when its length and 
breadth are given ? 



136 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

(2) How is one side found when the area and the other side 
are given ? 

43. What is the area of a field x -+- 1 rods long and x 3 rods 
wide? 

44. What is the area of a field whose length is x ft. and width 
| x ft. ? Of another field whose length is x + 10 ft. and width 
| x + 5 f t. ? 

45. What is the width of a field containing x acres whose 
length is 100 rods ? Whose length is a rods ? 

46. What is the width of a garden whose area is 165 sq. ft. 
and length 2 x - 9 ft. ? 

47. " Work and cistern J:> problems are solved by using recipro- 
cals. If A can do a piece of work in 7 days, what part of it can 
he do in 1 day ? In x days ? If a cistern can be filled by a pipe 
in x minutes, what part of it can be filled in 1 minute ? In 5 
minutes ? 

48. If a job of work can be done in x -f 1 da., what part of it 
can be done in 3 da. ? 

49. If a piece of work can be done by 1 man alone in 15 days 
and by another man alone in 20 days, how much can be done by 
both men in x days ? In x + 2 days ? 

50. If a cistern can be filled by one pipe in x minutes and 
emptied by another in x -f- 5 minutes, what part of the cistemf ul 
runs in in one minute if both pipes are open ? 

51. If a man can do a piece of work in 20 days, and a boy can 
do the same piece of work in x days, what part of the whole can 
both do in 5 days? 

52. If A can do a job in x days, B the same in | x days, and C 
in Jo; days, what part of the whole work can all working together 
do in one day ? 

53. Write down a number which when divided by x gives a 
quotient b and a remainder c. 



EXPRESSING SENTENCES AS EQUATIONS 137 

54. If a bill is shared equally among x laborers and each pays 
, how many dollars does the bill amount to? 

55. A man makes a journey of c miles. He travels x mi. an 
hour and a hr. by boat, x + 10 mi. an hour and b hr. by train, 
and the remainder on foot. How far does he travel on foot ? 

56. A floor is a ft. long and x + 2 ft. wide. How many 
square yards are there in it ? 

57. A horse eats x bu. and a cow b bu. of corn a week. How 
much will be left out of c bu. if they feed n weeks ? 

58. A boy is x yr. old and 5 yr. from now he will be half as 
old as his father. How old is his father now ? 

59. What is the interest on $1000 in b years at c% ? 

60. How much does a man save in a year if his salary is $# a 
month and he spends $b each month for 6 months, and $c a 
month for the remaining 6 months ? 

61. How long will one man require to mow a acres if x men 
mow & acres in a day ? 

62. How long will it take a person to walk b mi. if he walks 
20 mi. in x hr.? 

63. A man drives m hr. at the rate of x mi. an hour. How long 
will it take him to walk back at the rate of b mi. an hour ? 

64. A man bought b yd. of cloth at $c a yard and gained $#. 
What per cent did he gain ? 

92. Exercise in expressing Sentences as Equations. Two kinds 
of sentences may be distinguished : (1) those in which the 
equality of two described quantities is expressly stated; (2) those 
in which an equality is implied. 

1. The double of x is 24. Ans. 2^ = 24. 

2. Three times the sum of x and 4 exceeds b by 12. Ans. 
3(<B + 4) -6 = 12. 

3. Twenty subtracted from 2x gives the same number as 5 
added to a. 



138 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

In the problems on this and the next page, change the first 
sentence into algebraic language, letting x denote unknown num- 
ber. Answer second sentence by solving for x. Prove answers 

4. A man sold a horse for $140, gaining 12% of the cost. 
How much did the horse cost? 

SUGGESTION. x + ^-fax = 140. 

5. A lawyer took out 5 % commission from a bill he had col- 
lected, remitting $237.50. How much did he collect? 

6. A merchant sold a bedroom set at an advance of 35 % for 
$ 54. How much did it cost him ? 

7. A man spends $ 1239, which is 84 % of his salary. What 
is the number of dollars in his salary ? 

8. A sum of money (say $1), doubled itself at 5% simple 
interest in a certain number of years. What was this time ? A 
sum trebled itself at 7 % in a certain time. What was the time ? 

9. A house cost $4144, which was 40% more than a barn 
cost. What was the cost of the barn ? 

10. $860 yielded $247.25 at 5 % interest in a certain time. 
What was the time? $175.12 yielded $6.43 at 6%. In what time? 

11. The interest on $475 for 3 yr. 4 mo. at a certain rate was 
$ 95. What was the rate ? 

12. A certain sum which ran 8 mo. 15 da. bearing 6-L% interest 
amounted to $ 460.75. What was the principal ? 

13. $240.15 at 6% interest for a certain time amounted to 
$275.49. What was the length of time? 

14. B deposited 85 % of his money in a bank and then drew out 
20% of the sum deposited, leaving $3859. What sum had he? 

15. C bought hogs and sheep for $12,210, paying 65% more 
for the hogs than for the sheep. How much did the sheep cost? 

16. A town's taxable property of $1,360,000 at a certain rate 
yielded $8840 for school purposes. What was the rate? Mr. A 
in this town paid on his lot $78. What was its assessed value? 



PROBLEMS 139 

17. The population of New York state in 1905 was about 
8,069,700, and it had gained 11% from 1900. What was the 
population in 1900 in round numbers ? 

18. The amount of a certain principal at 5 % in 4-J- yr. is 
$ 796.25. What is the principal ? 

19. The distance from New York to San Francisco via Panama 
Canal is only 32 % of that via Cape Horn, and approximately 
10,098 mi. are saved. What is the length of each route ? 

20. The distance from Liverpool to San Francisco via Panama 
Canal is only 51 % of that via Cape Horn, and approximately 
7203 mi. are saved. What is the length of each route ? 

21. The proceeds of a note discounted at a bank at 6 % 2 mo. 
before it was due were $76.23. What was the face of the note? 
The proceeds of a sum discounted at 8 % 54 da. before due were 
$178. What was the sum ? 

22. A certain sum bearing 6 % compound interest amounted in 
3 yr. to $2977.54. What was the sum ? 

23. A draft on Chicago was bought at \/ discount for 
$ 6398.30. What was its face ? A draft on New York was bought 
at ^ % premium for $ 64.40. What was its face ? 

24. Find the value of a cargo insured at 3^ % for -^ of its value, 
if, upon its shipwreck, the company suffers a net loss of $8407.08. 

25. Including duty of 25 %, a man paid $918.75 for 420 yd. of 
broadcloth. What was the original price per yard ? 

26. A merchant sent his agent $ 14,616 to pay for cotton and 
agent's commission of 1^-%. How much cotton did he get? 

27. Coal bought at $6 a long ton (2240 lb.), was sold by short 
ton (2000 lb.) at same price. What per cent was gained ? 

28. When a railroad stock yields 4 % and is 20 % below par, 
a certain sum invested in it produces $1200 income. What sum ? 
When stock yields 5%, is 12% above par and produces $210? 

29. 20,672 lb. hams sold; commission 2i% ; guaranty 2|% ; 
net proceeds due consignor $ 2448.34. What was price per lb. ? 



140 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

30. A fish weighing x Ib. was caught whose tail weighed 4 Ib. ; 
its head weighed \ as much as the whole fish, and its body J- as 
much as the whole fish and 3 Ib. more. 

31. A man divides $1.35 among 12 children, giving x of them 
18 ^ apiece, and the remainder 12 x apiece. 

32. A sum of money amounting to $x is divided up, A receiv- 
ing | of it, B ^, C -|, and D the remainder, $8. 

33. A piece of work is performed by A doing x sixths, B doing 
x ninths, and C doing x twelfths. Whole work is represented by 1. 

93. Equations changed to Sentences. To throw more light on 
the changing of sentences into equations, a short exercise is given 
on the converse operation. 

Ju JU -| p* 

1. x - = - + 15. 

Ans. The difference between a number and the third part of it 
is equal to the fourth part of it increased by 15. 

o x x 1 2 K x . x x 

2. = 



7x + 5 3x + a 2x-b 

o. - = - . D. - = - . 7. - = - 

32 28 c d 

8. If C's age is x, B's is 3 x, and A's 6 x, change 



into a sentence, stating how many times C's age, B's and A's ages 
are respectively. 

9. If Charles has $o?, John $6 a;, Henry $5#, change 

x + 6 x +5# = 54 

to a sentence, stating how many times greater John's and Henry's 
money each is than Charles's. 

10. Change x + (x -f- 1) + (a; + 2) + (x + 3) = 65 to a sentence. 
SUGGESTION. These numbers can be called consecutive numbers. 



PROBLEMS 141 

94. The Solution of Problems involving One Unknown Number. 

The solution of a problem consists of two parts : 

A. The statement or stating of the problem, that is, the getting 
of the equation. 

B. The solution of the equation found. 

We will now present more fully all the steps in the solution of 
a problem : 

(1) Reading the problem, understanding its meaning, and getting 

all the different conditions in mind. 

(2) Looking for the unknown number and letting x, orn, etc., equal it. 

(3) Constructing the f auctions of x (one or more) described in the 

problem. 

(4) Forming the equation by turning a sentence in the given prob- 

lem into algebraic language. 

(5) Solving the equation obtained by the method of 87. 
(G) Proving the answer or answers correct. 

a. If there are two or more unknowns described in the problem and x is 
used to denote one of them, then the others are denoted by functions of x. 

b. It often happens that it is convenient to use a function ofx instead of 
#, to denote the unknown number, or the first of two or more unknown 
numbers. Suggestions will be made when it is wise to do this. 

95. Ratio and Proportion Problems. The language of ratio and 
proportion is often used in the wording of problems. Proportions 
can always be changed into equations by setting the product of 
the first and last or extreme terms of the proportion equal to the 
product of the two middle terms or means. 

96. Meaning of " = " in Algebra. The sign = is used, one time 
and another, in a considerable number of different senses. Thus, 
in arithmetic, we may say 2 yards = 6 feet, the sign here mean- 
ing " has the same length as " ; 1 ton coal = $ 3.20, sign now 
meaning " costs " ; area of rectangle = 20 square meters, the sign 
meaning " covers same surface as," etc., etc. In algebra, however, 



142 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

= always has one meaning, namely, that the number on one side 
is the same as the number on the other side. Hence, in algebra we 
cannot express one side of an equation in cents and the other in 
dollars j nor one side in yards and the other in feet, etc. 

PRQBLEMS 

97. Problems containing One Unknown Quantity. 
1. What number is that whose half, third, and 100 more equals 
twice the number and 2 more ? 

SOLUTION. Let x = the number. 



Then, - = one half of the number. 
2 

f = one third of the number. 
2 x + 2 = twice the number and 2 more. 



Functions. 



| + - + 100 = 2 x + 2. (Sentence to Equation.) 

I. > 

Sx + 2x + 600 = 12z+12. (Mult. Ax.) 

_ 1x = _ 588. (Sub. Ax. and Collecting.) 

x = 84. (Div. Ax.) 

VERIFICATION. Half of 84 is 42 ; one third of 84 is 28 ; 42 + 28 + 100 = 
170 ; also twice 84 and 2 more is 170. 

a. Do not verify problems by substituting the answer in the equation as 
was the rule in 87, since a mistake may have been made in getting the equa- 
tion. Instead, test the answer by seeing whether it satisfies the reading of 
the problem. The student should make it his regular practice to test all 
answers in this way. 

2. What number is that whose half, third, and fourth parts 
together equal 65 ? 

3. What number is that which being increased by its six 
sevenths and the sum diminished by 20 equals the result 45 ? 

4. What number is that to which if 1H be added, 4 times the 
sum will be equal to 10 times the sum of the number and one ? 



PROBLEMS 143 

5. A certain number multiplied by 5, 24 subtracted from the 
product, the remainder divided by 6, and this quotient increased 
by 13 results in the number itself. What is the number ? 

6. Demochares has lived a fourth of his life as a boy; a fifth 
as a youth ; a third as a man ; and has spent 13 yr. in his do- 
tage : how old is he ? (From a collection of questions by Metro- 
dorus, 310 A.D.) 

7. Divide the number 181 into two such parts that 4 times 
the greater exceeds 5 times the less by 49. 

SUGGESTION. Here there are two unknown numbers. Let x = the greater 
number. What then is the lesser number, if their sum is 181 ? 

8. "Heap, its seventh, its whole, it makes 19," that is, find a 
number such that the sum of the number and one seventh of the 
number equals 19. (From Ahmes Collection of Problems, made 
in Egypt 1700 (?) B.C.) 

9. What part of the day has disappeared if the time left is 
twice two thirds of the time passed away ? (Palatine Anthology, 

300 A.B.) 

SUGGESTION. Let x = number of hours of the 24 that have passed away. 
Then how many remain ? 

10. Diophantus of Alexandria, the reputed first writer on 
algebra, had the following engraved on his tombstone : Diophantus 
passed i of his time in childhood, yL in youth, and 1 as a bachelor ; 
5 years after his marriage was born to him a son who died 4 
years before his father at half his father's age when he died. 
How long did Diophantus live ? 

SUGGESTION. Let x = number of years D. lived. Then the sum of all the 
periods named (including the age of his son) =? 

11. The British House of Commons in 1908 had 670 members. 
Ireland had 5 less than three halves of Scotland's number and 

A 



144 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

England had 20 less than 5 times Ireland's number. How many 
members came from each country ? 

12. A person at the time of his marriage was three times as old 
as his wife, but 20 years later he was only twice as old. What 
were their ages on their wedding day ? See 73, Ex. 22. 

13. A father aged 54 years has a son aged 9 years; in how 
many years will the age of the father be just 4 times that of the 
son? 

14. A father is now 40 years of age and his daughter 13. How 
many years ago was the father's age 10 times that of the daugh- 
ter? 

15. The Pennsylvania Railroad from New York to Chicago is 
912 miles long. From Philadelphia to Pittsburg is 6 miles less 
than 4 times the distance from New York to Philadelphia, and 
from Pittsburg to Chicago is 24 miles more than from New York 
to Pittsburg. What are the distances from New York to Phila- 
delphia and from New York to Pittsburg; also the distance from 
Pittsburg to Chicago ? 

16. The difference between two numbers is 12 and the greater 
is to the less as 11 : 5. What are the numbers ? 

SUGGESTION. Let llcc=the greater number and 5x the less number. 
Follow a like course in solving Ex. 17-19. See 94, b. 

17. In Georgia in 1900 there were not far from 8 white persons 
to 7 negroes. The population of the state was very nearly 
2,216,100. How many were whites and how many negroes ? 

18. In Tennessee in 1900 there were not far from 16 whites to 
5 negroes and the population of the state was very nearly 
2,020,410. How many were whites and how many negroes ? 

19. In 1900 in the United States for every 104 persons engaged 
in agriculture there were very nearly 71 engaged in manufactures, 
57 in domestic service, 48 in trade and transportation, and 13 in 
professional pursuits. If the same ratio holds and the total popu- 
lation is taken, as estimated by governors of the states Jan. 1, 



PROBLEMS 



145 



1908, to be 88,912,022, how many persons were engaged in each 
form of industry ? 

20. J. Nilsson made a skating record February, 1895. In Feb- 
ruary, 1887, on a straightaway course with the wind at his back, 
Tim Donoghue made the mile in 23.4 seconds less time than 
Nilsson. Nilsson's time was to Donoghue's as 20 is to 17. What 
were the two times ? 

21. The needle of a talking machine follows a spiral (see 
figure) ; suppose we think of the 

spiral as five circumferences whose 
sum is 75.57 in., the smallest equal- 
ing 15 in. What is the common 
distance between them if it is the 
common difference between two ad- 
jacent circumferences, x, divided by 

6f(27r)? 

22. A vote was taken in a debat- 
ing society arid the motion was car- 
ried 5 to 3 ; on reconsideration 110 
affirmative votes deserted to the 

negative and the motion was lost 3 to 4. What was the number 
of affirmative and the number of negative votes on first ballot? 

23= The number of representatives in Congress by the appor- 
tionment of 1900 was 386. Of this number the Southern had 5 less 
than 6 times as many as the Western states ; the North Central 
had 6 less than Western and Southern together, and the Eastern 
28 less than the North Central. How many had each region? 

24. The area of New York state is 8164 sq. mi. more than that 
of Ohio, and Texas has 4836 sq. mi. more than 3 times New 
York's and Ohio's areas together. The area of all three is 
356,140 sq. mi. What is the area of each state ? 

25. Mt. Shasta is 203 ft. higher than Pike's Peak, and Mt. 
McKinley is 6114 ft. higher than Mt. Shasta. The sum of the 
three heights is 48,961 ft. What is the height of each? 




146 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

26. In the National League in 1905 Pittsburg won 9 games 
less than New York, Chicago 4 less than Pittsburg, Philadelphia 
9 less than Chicago, Cincinnati 4 less than Philadelphia, St. 
Louis 21 less than Cincinnati, Boston 7 less than St. Louis, and 
Brooklyn 3 less than Boston. In all 612 games were played. 
How many games did each team, win ? 

27. A rectangular field is 6 rods longer than it is wide ; and if 
the length and breadth were each 4 rods more, the area would be 
120 square rods more than it is. Find the length and breadth of 
the field. 

28. The length of a rectangular floor exceeds the width by 6 
ft. If the width is increased by 3 ft. and the length by 2 ft., 
the area is increased by 134 sq. ft. Find the length, breadth, 
and area of the floor. 

29. The length of a room exceeds its breadth by 8 ft. If each 
had been increased by 2 ft., the area would be increased by 60 sq. 
ft. Find the original dimensions of the room. 

30. The total number of persons speaking English, German, 
and French in 1801 has been estimated at 82 millions. Of this 
number the German language was spoken by 9 millions more than 
the English, and the French by 1 million more than the German. 
How many millions spoke each language ? 

31. The total number of persons speaking English, German, and 
French in 1901 was 266 millions. Of this number the English 
language was spoken by 46 millions more than the German, and 
the German by 32 millions more than the French. How many 
spoke each language at this time? 

32. A man invested $1600, a part at 6 % and the rest at 5 %. 
If his total annual income was $ 90, how much did he invest at 
each rate ? 

IMPORTANT NOTE. Heretofore the unknown number has usually been 
denoted by x. The student should now accustom himself to using other 
letters instead, preferably initials. Thus, use n (number) in Ex. 32-35, 38 ; 
r in 36 j v (velocity) in 37 ; / (franc) in 39 ; and so on. 



PROBLEMS 147 

33. A man has two thirds of his property invested at 4 <f , one 
fourth at 3%, and the remainder at 2%. How much property 
has he if his annual income is $1720 ? 

34. A man put out $12,990 in two investments. On one of 
these he gained 12%, and on the other he lost 5%. If his net 
gain was $ 753, what was the amount of each investment ? 

35. A person who is worth $12,000 invests a certain amount 
in railroad stock from which he receives 6%. One third of the 
remainder pays 4% interest, and the other two thirds 5%, and 
from all his funds he clears $ 600. What number of dollars was 
invested in railroad stock ? 

36. A man has $ 5000 invested at a certain rate, $ 2000 at 
double the former rate, and $ 1000 at three times the first rate. 
His income from these funds is $GOO. What rate of interest 
does he receive 011 the $ 2000 ? 

37. A just perceptible breeze has one tenth, and a gentle breeze 
one fourth of the velocity of a brisk wind. A storm has 21 times 
and a hurricane 4J times the velocity of a brisk wind. Now if 
the first four velocities are added, they fall 13 nii. per hour short 
of the velocity of a hurricane. What is the velocity of each ? 

38. The silt and rock waste carried along by the lower Missis- 
sippi every year amounts to 8848 million cu. ft. Of this sum the 
silt dragged along the bottom is 50 million cu. ft. more than half 
the number of cubic feet of minerals carried, and the silt carried 
in suspension is 3 million cu. ft. less than 5 times the number of 
cubic feet of minerals. How many millions of cubic feet are 
there of each ? 

39. A French franc, a German mark, and an English shilling 
are worth together 67.4 cents. A shilling is worth 5 cents more 
than a franc, and a mark half a cent less than a shilling. What 
is the value of each in United States money ? 

40. A can perform a piece of work in 6 da., B can perform 
the same work in 8 da., and C in 24 da. In how many days 
can they finish it if all work together? 



148 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

SUGGESTION. "Work" problems, "cistern" problems, and other prob- 
lems like them are solved by finding the part of the whole that is completed 
in one day (or hour, or minute). 

Let n = the number of days it will take all to complete the work. What 
part of the work will A do in one day ? In n days ? B in n days ? C ? All ? 

What then is the equation ? Ans. + -+;- =!. Ans. to problem, n = 3. 

VERIFICATION. If A works 3 days, he does f or half the work ; B does f , 
and C does ^ or 1. Now, \ + | + \ = 1, or the whole job. 

41. A can do a piece of work in 10 da., and B can do it in 
15 da. How long will it take both? 

42. A can do a piece of work in 10 da., B in 12 da., and C 
in 8 da. In how many days can all working together do it ? 

43. A can pave a walk in 6 da., and B in 8 da. How long 
will it take A to finish the job after both have worked 3 days ? 

44. Three pipes are connected with a cistern. It takes the 
first pipe 20 min. to fill it, it takes the second pipe 30 min. to 
fill it, and it takes the third pipe 60 min. to empty it, the last 
pipe carrying out instead of in. If all the pipes are left open, 
how long will it take to fill the cistern ? 

45. A can build a wall in 15 da., but with the aid of B and C 
the wall can be built in 6 da. If B does as much work in 1 da. 
as C does in 2 da., in how many days can B and C separately 
build the wall ? 

46. A tank is emptied by two pipes. One can empty the 
tank in 30 min. ; the other in 25 min. If the tank is f full, and 
both pipes are opened, in what time will it be emptied ? 

47. There are two faucets in a kitchen ; a bucket can be filled 
in 3 min. from the hot water faucet, and in 70 sec. from the cold 
water faucet. How long will it take to fill the bucket if both 
faucets run with the same pressure as when they are opened 
singly ? 

48. Four pipes discharge into a cistern ; one fills it in one day ; 
another in two days ; the third in three days ; the fourth in four 



PROBLEMS 149 

days ; if all rim together, how soon will they fill the cistern ? 
(Collection of Metrodorus.) 

49. Six dollars is changed into 51 coins of two kinds, quarters 
and dimes. How many were there of each ? 

50. Five boys bought a boat, agreeing to share the expense 
equally. But two of them having left $ 1 of his share unpaid, 
each of the others had to pay one twelfth more than his share. 
What was the cost of the boat ? 

*51. It is estimated that there are 1440 millions of inhabitants 
on the earth. Of this number there are 20 millions more yellow 
than white persons ; 4 times the number of blacks and 10 millions 
more equals the number of white persons ; 40 times the number 
of red persons and 10 millions more equals the number of white 
persons; and 2- times the number of red men equals the number 
of brown men (Malays and Polynesians). How many are there 
in each race ? 

52. The population of New York City in 1776 was 461 more 
than 20 times what it was in 1653; in 1860 it was 9327 less 
than 36 times what it was in 1776 ; in June, 1908, it was estimated 
to be 269,977 more than 5 times what it was in 1860. Now the 
population in 1908 was 17,593 more than 189 times what it was 
in 1776. Find the population at each date named. 

53. A boy buys apples at the rate of 5 for 2f. He sells 
half of them at the rate of 2 for 1 f and the rest at the rate of 3 
for 1 $, and clears 1 1 by the transaction. How many did he 
buy? 

SUGGESTION. Let n = the number of apples he bought. What was the 
price of the apples, or cost of one ? Then what was the total cost ? Next, 
what was the selling price of one apple of the first kind ? How many 
were sold at this price ? How much did they amount to ? What was the 
selling price of one of the second kind, etc. ? 

54. A boy bought apples at the rate of 3 for 1 ^. He sold two 
thirds of them at the rate of 2 for 1 ^, and the remainder at a 
cent apiece, and cleared 24^. How many did he buy? 



150 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

55. I bought a certain number of apples at 4 for a cent. I 
kept one fifth of them, and sold the rest at 3 for a cent, and 
gained 1 cent. How many did I buy ? 

56. A bought eggs at 18 J^ a dozen. Had he bought 5 more for 
the same money, they would have cost him 2-J-^ a dozen less. 
How many eggs did he buy ? 

57. The immigration into the United States in the year 1870 
was 236,966 more than it was in 1860; in 1880 it was 70,054 
more than it was in the year 1870 ; in 1890 it was 1955 less than 
in 1880; in 1900 it was 6730 less than in 1890 ; while in 1907 it 
was 66,784 less than 9 times what it was in 1860. Now the total 
immigration for the four years, 1860, 1870, 1880, 1890, was only 
164,650 more than for the single year 1907. What was the 
immigration in each year named ? 

58. A train runs 200 mi. in a certain time. If it were to run 
5 mi. an hour faster, it would run 40 mi. further in the same time. 
What is the rate of the train? 

59. A train runs 84 mi. in the same time a second train runs 
96 mi. If the rate of the first train is 3 mi. per hour less than 
that of the second train, find the rate of each. 

a. In every such problem after letting r = the unknown, it will be found 
that counting r or some function of r, as given, two of the three, rate, time, 
and distance are given to find the third. When the student has constructed 
the third as a function of r, he will usually find himself well along towards 
getting the equation. 

60. A train on the Northwestern Line passes from London to 
Birmingham in 3 hr. ; a train on the Great Western Line, which 
is 15 mi. longer, traveling at a speed which is less by 1 mi. per 
hour, passes from one place to the other in 3 l j hr. Find the 
length of each line. 

61. A and B are 240 mi. apart. If two trains leave A and B 
at the same time for the other place as destination, the first 
traveling 45 mi. an hr. and the other 35 mi. an hr., how far from 
A will they meet ? 



PROBLEMS 151 

62. A leaves New York and travels at the rate of 11 mi. in 5 
nr. ; 8 hr. later B leaves New York following A's route at the 
rate of 13 mi. in 3 hr. How far must B go to overtake A ? 

63. A person walks to the top of a mountain at the rate of 21 
mi. an hour, and down the same way at the rate of 3.V mi. an hour, 
and is out 5 hr. How far is it to the top of the mountain ? 

64. In one hour an automobile runs 10 mi. farther than a man 
on a bicycle, and his rate is 8 mi. an hour. If it takes the auto- 
mobile 5 hr. less time to run a certain distance than it takes the 
man to ride it, what is the distance ? 

65. At the rate of 3 mi. an hour uphill and 4 mi. an hour 
downhill, a man can walk 60 mi. in 17 hr. How much is 
uphill and how much is downhill, if the distance has no part 
level? 

66. A steamer's speed is such that on a certain stream it takes 
as long to go 3 mi. upstream as to go 5 mi. downstream, and the 
rate of the current is 2 mi. an hour. AVhat is its rate in still 
water ? 

67. A boatman, who can row 6 mi. an hour in still water, rows 
a certain distance downstream and back again in 3 hr. How far 
did he row, supposing the stream flowed 3 mi. in 2 hr. ? 

68. The distance from. Albany to Syracuse is 148 mi. A canal 
boat leaves Albany for Syracuse, moving at the rate of 3 mi. in 
2 hr. ; another boat leaves Syracuse for Albany at the same time, 
moving at the rate of 5 mi. in 4 hr. How far from Albany do 
they meet ? 

69. From Chicago to Denver over the Burlington is 1018 mi. 
Using Central time only, train No. 1 leaves Chicago (winter 
schedule, 1908) at 1 P.M. and travels 34 mi. an hour ; train 
No. 2 leaves Denver at 9 P.M. of the same day and travels 31.3 
mi. an hour. In how many hours from the time No. 1 leaves 
Chicago will they meet ? Also how far from Chicago will they 
meet ? 



152 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 

70. In a mile run B has a handicap over A of 80 ft., that is, he 
starts at the same time A does but 80 ft. in front of him. A runs 
15 ft. while B runs 14, and A wins by ^ minute. Required, A's 
time to run the mile in minutes. 

71. The distance from M to L is 311 m i. The express down 
train leaves M at 11.30 A.M., and arrives in L at 12.30 P.M. The 
up train leaves L at 11.45 A.M., and arrives at M at 12.35 P.M. 
If the speed of both trains is uniform, when will they meet ? 

n x rate down train. P (n 15) x rate up train. 

SUGGESTION. First get the rate of speed of each train from the whole 
distance, and its corresponding time. Then let n number of minutes after 
11.30 when they meet. How many minutes then will the up train run until 
they meet ? It will be found that 31 \ can be divided out of the equation. 

72. If 65 lb. of sea water contain 2 Ib. of salt, how much 
fresh water must be added to these 65 lb. in order that the quan- 
tity of salt contained in 25 lb. of the new mixture shall be 
reduced to one fourth pound ? 

73. In an alloy of 90 oz. of silver and copper there are 6 oz. 
of silver. How much copper must be added so that 10 oz. of the 
new alloy may contain two fifths of an ounce of silver ? 

74. How many quarts of water must be mixed with 250 quarts 
of alcohol 80 % pure to make a mixture 75 % pure ? 

75. An officer attempting to arrange his men in a solid square 
found that with a certain number on a side he had 34 men over, 
but with 1 man more on a side, he needed 35 men to complete the 
square. How many men had he ? 

SUGGESTION. Instead of letting n = the number of men he had, we let 
n = the number of men on one side of the first square. 

76. A general who has his men arranged in a solid square finds 
after a battle he has lost 200 of them. Attempting to make a 
square with 1 less man on each side, he finds he has 21 men over. 
How many men were there in the first square ? 



PROBLEMS 153 

77. According to the United States census for 1900 there were in 
round numbers 14,794,000 males over 30 years of age. Twice the 
number of those having a college or university education and 5 
thousand more equals the number of those having finished a high 
school course or its equivalent. Eighteen times the number with 
a high school training and 228 thousand besides equals the num- 
ber of those who received a common school education. Also one 
seventh of those having a common school education is 35 thousand 
less than the number without education. How many were in each 
class ? 

78. Wlio's Who in America (edition 1903) contains the names 
of 11,384 persons of eminence. To show the immense advan- 
tage an education gives a person, a table has been constructed 
giving in per cents (1) the number in the book finishing a college 
course ; (2) the number taking part of a college course ; (3) the 
number finishing in a high school or academy ; (4) the number 
having a common school education. After dropping the 444 
persons privately educated, there remain 10,940 persons in the 
four classes named. Now 575 %, 150 %, 161.5 %, respectively, of 
the number in class (4) equal the numbers in classes (1), (2), (3). 
How many persons were in each class? 

REMARK. Get the answers to nearest integers. 

79. A merchant maintained himself for three years at a cost of 
$ 1250 a year, and in each of these years increased that part of 
his stock which was not spent by ^ of itself. At the end of the 
third year his original stock was doubled. What was the value of 
his original stock ? Ans. $ 18,500. 



CHAPTER V 

SIMULTANEOUS EQUATIONS CONTAINING TWO OR MORE 

UNKNOWN QUANTITIES* 

98. Simultaneous Equations are sets of equations whose unknown 
quantities have the same values. 

They are solved by a process called elimination. Elimination 
is a process of deriving from two or more independent equa- 
tions one equation containing but one unknown quantity. Liter- 
ally the word means turning out of doors (from limen, threshold, 
and e, out of). The unknowns are turned out of the equations. 

There are several recognized methods of elimination. The two 
most important are (1) elimination by substitution) (2) elimination 
by addition and subtraction. 

99. Elimination by Substitution is effected by finding the value 
of one unknown from one of the given equations and substituting 
this value for the same unknown in the other equation. 



1. Given the equations K 1 ) 2 x + 3 y =13 >\ to find values 

4 




of x and y that will satisfy both equations ( 82). 

a. Observe the equations are numbered (1) and (2). We need to mark 
the equations in some such way so that we can refer to them readily. 
Whenever, in what follows, either of the given equations has its form changed, 
it will no longer be marked (1) or (2), but these figures will have what are 
called subscripts, that is, small figures written under and after them. Thus. 
(li) means eq. (1) has been changed in form; (1 2 ) means (li) has been 
changed ; and so on. 

SOLUTION. (1) 2 x + 3 y = 13. 

(Sub. Ax.) 

(Div. Ax.) 

JU 

* If so desired , Chapter VI can be taken up before Chapter V. 

154 



ELIMINATION BY SUBSTITUTION 



155 



We now substitute this value of z, , in place of x in the second 

a 

equation. (See definition of simultaneous equations, p. 158.) Thus, 

(2) 3 x + 4 y = 18 becomes 3 / 13 ~ 3 ?/\ + 4 y = 18. 

\ 2 / 

,-. SJL^^J _|_ 4 y = 18. (Multiplication of fractions.) 

tt 

39-9?/ + 8?/ = 36. (Ax.?) 

-y=-S. (Ax. ?) 

y = 3. Ans. (Ax.?) 

To find the value of x, we substitute the value of ?/ just found in (12), the 
most convenient to use of the equations which contain both x and y. Thus, 
putting 3 for y, we have, 

13 ~ 3 x 3 



x = 



= 2. 



VERIFICATION. (1)2x2 + 3x3 = 13; (2)3x2 + 4x3 = 18. 



(2) 



< 

' l2a; 

f 12 



5 a? - 3 (a? - y) = 1 3. 



12 a? - - 1 7 = 4. 



(2) 



156 



SIMULTANEOUS EQUATIONS 



SUGGESTION. First, we clear each equation of fractions. Next, taking the 
second equation, we transpose so that the y terms will be on the left side and all 
the other terms on the right side, and then the value of y is easily found in 
terms of x, that is, the value of y contains x. We now substitute this value of 
y in the first equation, getting an equation containing only one unknown, like 
those in 87. Solving this equation, we substitute the value of x found in 
one of the given equations to get the value of y. Ans. x = |, y = . 



15. 



2' 



16. 



13 7 






S+21.1T. 



b. The student has probably observed that we can solve either equation 
for either unknown and substitute the value found in the other equation. 
As a rule we take the simpler equation and solve it for the unknown with 
the smaller coefficient. It is usually best to have the other equation in its 
simplest form before substituting. 



17. 



19. 



'x -\- 



x 

T 



x + y x y 



18. 



x + y 



=1-* 



2 



= 8, 



20. 



i 



4 
3ft 2 __3j/ + 7 



x 



21. 



I 2 a? + 5 y = 35. 



1 5 ft - 33 y = 0. 



23. Make a rule for solving simultaneous .equations containing 
two unknowns, using elimination by substitution. 



: 24. 



3ff+4?/ + 3 2ft + 7-?/__ 5 ?/--8 

10 15 5 ' 

9 7/4. 5 ft --8 ft + ?/ _ 7 ft + 6 

12 4 11 "' 



ELIMINATION BY ADDITION AND SUBTRACTION 157 



25. 



26. 



7 + x 2x-y = _ 3 - 5?/- ^ 4a;-3 J8 






1.5 



4.2 



-.7. .5z-^2 41 
"5 



1.6 



27. 



6# + 9 3 a; + 5 y = 
4 4 a; 6 



16 



+ 4 



10 



7 6a?-3y 
" - 4) 



-4 = 



100. Elimination by Addition and Subtraction. 

1. Given (1) 2x - 9 y = 11 ; (2) 3 a; - 12 y = 15. 

SOL. (2i) 6 x - 24 y = 30 (Mult. Ax.) 

(li) 6 x - 27 y = 33 (Mult. Ax.) 

3 y = - 3 (Sub. Ax. Explain.) 

2/ = 1, vlns. (Div. Ax.) 

(1) 2 x - 9(- 1) = 11 ; or 2 x + 9 = 11 ; whence x = 1. .4n*. 



VERIFICATION : 



(1)2x1- 9(-l) =11, or 2+ 9 = 11; 
(2) 3 x 1 - 12(- 1) = 15, or 3 + 12 = 15. 



EXPLANATION. Observe that equations (1) and (2) were multiplied 
through by such numbers as made the coefficients of x the same in the two 
equations. If these coefficients (as and 6 in (2i) and (li) above) have like 
signs, one equation is subtracted from the other to eliminate the unknown. 
If the two arithmetically equal coefficients have unlike signs, one equation is 
added to the other. Strictly speaking, it is not the equations that are added, 
or subtracted, but only their corresponding members, though this language 
is often used. 

Evidently if it is more convenient to do so, instead of making the coeffi- 
cients of x the same, one can make the coefficients of y the same. 



2. 



f4+9y=61, 



> 

I Qx y = l. 



6. 



4. 



7. 



158 SIMULTANEOUS EQUATIONS 

g flloj+6y=64, 9 | 7 a?- 9^ = 7, 1Q f 1.5a?-3.7y=-56, 
il3a;+102/=96. ' la?-2y = 3. 

u . (i) +2y + 



SOLUTION 



12- 



(11) 9a;+122/-6a: + 4y = 252 (2i) 45- 20^=000+3 a; (Mult. Ax.) 

(1 2 ) ax + 16y=262 (2 2 ) 42 a;- 20?/ = 600 (Ax.?) 

(1 3 ) 42 a; + 224 y= 3528 (Ax. ?) 

244^=2928 (Sub. Ax.) 

?/ = 12 (Ax. ?) 
(1 2 ) 3 x + 192 = 252, whence x = 20. 

VERIFICATION. (1) %- + 24 - 20 + - 2 /-=42, (2) GO - -^ = 40 + ^>-. 

3a?-5y__2a; + 4 , 

- 



17. Make a rule for eliminating by addition and subtraction by 
answering the following questions : 

(1) If the given equations contain fractions, what is done ? 

(2) When is transposing necessary and how is it done ? 

(3) How is the unknown to be eliminated selected ? 

(4) What is the coefficient of the unknown to be eliminated made 

in each of the next pair of equations ? Ans. The least 
common multiple of the two coefficients of this unknown 
in the two equations just found. 



ELIMINATION BY ADDITION AND SUBTRACTION 159 

(5) How is the number by which to multiply the first equation 

found ? The number by which to multiply second equation ? 

(6) After the coefficients of either x or y are made the same in the 

two equations, what is done if these coefficients have the 
same sign ? If they have opposite signs ? 

(7) After the value of one unknown is found, how is the other 

found ? 

(8) In what equations is the verification performed ? 



18. -13 = 






19. = _; 9a; _ = 

x-y 8 7 



20. = . 

? 14 



2 i. ii 



O ! 7 O / 

11 -# , 4 a; -f 8?/ 2 



22 -- = . --_. 5 

6 3 "G 3 6 



23. ^-20- = - 

' 



'23 -x 2 a? -18 3 



x+5 6x-2. 2y-3 , . . 

" -~~' ~~ + = ~~' ( 



25. (1) - = 2; (2) + = . 

; x y ' ^ xy 3 

In the solution which follows we treat - and - ('called the reciprocals of x 

x y 
and / as the unknowns. 



160 



SIMULTANEOUS EQUATIONS 



(li) -- =6 (Mult. Ax.) 
x y 

(2i) + = (Mult. Ax. ) 
x y 3 



29 58 , c , . , 

- = - (Sub. Ax.) 

y 3 

87 = 58 y. (Mult. Ax.) 



VERIFICATION. 



58 2 

(1) - = 2, whence x = 1. 

x 4 



0) ?= 



26. 



28. 



30. 



1 + 1 - 1 

x + ^"2' 



.05 



14 

0? 



L = l 
/~6' 

25 = 7 



49 _50__ o 



. x 



y 



a; 
9 


y 

22 


8 
= 15* 

_ 3 

Z 25 


?/ 



27. 



9 



5 

y 

10 



-- = 1. 



*29. 



[21.1 


, 12 1 

-p -iw = 


5 


X 


y 




1 


i i 






"#~~42* 





31. 



2 ?/ x = 4 #?/. 



101. Elimination by Comparison is performed by finding the 
value of the same unknown from each of the two given equations 
and setting these values equal to each other. (See Ax. 7.) This 
method is rarely better than one or other of the two methods just 
learned, and exercises on it are not given. 

102. Miscellaneous Exercise in Elimination. The choice of the 
method in particular problems is left to the student, but he should 
not confine himself to one. 



EXERCISE IN ELIMINATION 



161 



1. 



3. 



5. 



/ J fl 


o; 


1 Jl 


Y 


3 + 5 ' 


4 


' (> 


2* 


2. 








/ O* ?/ 

~ i~\ 


Ox' 


y 


1 


I 3 5~ ' 


[3 


8 


2 


x 4 y 4- 4 


r u 


+ 3 


5 


a,. _3 y + 7' 


V 


+ 3 


~6' 


4. 








. l; + 5_ 2 /-l 


u 


-3 


2 


U + 2 y-2 


.V 


-3 


3 


(2x 3y 00 


r 


5 


7 


3 4 


X 


4-2 


y 2 x + y 


6. 








y _|_ 1 00 




7 


5 


^/ 1 J-V/V7 


.3x-2 Gy-7 



7. 



4 ^ + 5 ?/ 25 



= 3, 



8 x + ?/ 4- 6 
-L^- - = o. 



8. 



23 -a? 



4- - 



a; -18 



= 



73 -3 



( 88.) 



9. 



2y 5-z^41 2y l 
3 2 : 12" 4 



*10. 



11. ^ 



2a; + 8 



10 



15 



5rc--15 a; y + 1 _ 7 a; 4- 12 
12 3 11 



12, 



18 



x y 4 



r=l( ;+: : J+iV ( 100 ' 25 > 
a? ?/ oVa; ?/y 12 



162 SIMULTANEOUS EQUATIONS 

103. Special Methods of solving Simultaneous Equations. Systems 
of Equations. Instead of eliminating immediately, it is often 
better to derive one or more new equations from the given ones, 
and let such equation or equations take the place of one or more 
of the given ones. 

1. (1) 9 a + 130 = 184, (2) 13^ + 190 = 268. 

SOLUTION. (h) 27x + 39y = 552 (Ax. ?) 

(2i) 26 x + 38 y = 536 (Ax. ?) 

(3) x+ y= 1(J (Ax. ?) 

(3i) 9 x + 9 y == 144 (Ax. ?) 
(1) 9a+13y== 184 

4y- 40 (Ax. ?) 

y= 10 (Ax. ?) 

(3) +10= 16;.-. X = 6. 

VERIFICATION. 9 x 6 + 13 x 10 = 184; 13 x 6 + 19 x 10 = 268. 

2. (1) 140-170 = 159; (2) 29 x ~ 37 y = 324. 
SUGGESTION. Multiply (1) by 2 and subtract from (2). 

*3. (1) 1390+152^=1377; (2) 35 x + 37 y = 348. 

4. (1) 755 x - 564 # = 2074 ; (2) 1133 x- 847 y = 3113. 

5. (1) x+ 500 = 557; (2) 500 + = 361. 

SUGGESTION. First add the two equations and divide by 51 ; then subtract 
(2) from (1) and divide by 49. 

a. In each of the preceding problems (1) and (2) constituted the given 
"system of equations." In exercises 1, 2, 4, the given system is replaced 
by the system composed of equations (1) and (3). In exercise 3, it is better 
to replace the given system by (2) and (3). In exercise 6, the given 
system is replaced by (3) and (4). The new systems in each case have the 
same values of x and y because they were obtained by reversible axiomatic 
processes. These new systems are formed because in such problems as the 
preceding, they greatly shorten the labor of calculation. Thus, by actual 
count, to solve exercise 2 by regular process would require 99 figures ; the 
above method calls for only 46 ; besides, the labor is easier to perform. 

6. (1) 590 + 730 = 390$; (2) 73 x + 50 // = 379J. 

7. (1) 23 x - 41 y = - 2y, ; (2) 39 - 2,1 y = 6|. 

8. (1) 2S0-350 = r>G; (2) 29 -x- 1, //= 151. 



PROBLEMS --TWO UNKNOWN NUMBERS 163 

104. Problems involving Two Unknown Numbers. The sugges- 
tions given in 94 apply equally here. 

1. A grocer has two kinds of sugar, one worth 5 4 and the 
other 6^ a pound. How many pounds of each sort must be 
taken to make a mixture of 100 Ib. worth 5J^ a pound? 

SOLUTION. Let x no. of Ib. of 5 ^ sugar, 
and y = no. of Ib. of 6 ^ sugar ; 

then 5 x no. of cents in cost of 5 ^ sugar, 

and 6 y no. of cents in cost of 6 / sugar. 

Now by the conditions of the problem we have two equations : 

(l)x+y=lQQ. (2) 5x + 6ij = 575 

(li) 5 x + 5 y = 500 (Ax.?) 
y= 75 (Ax.?) 
(1) .-.x= 25 

VERIFICATION. 5 x 25 + 6 x 75 = 100 x 5f ; also 25 + 75 = 100. 

2. Find two numbers whose sum is 196 and whose difference 
is 8. 

3. In a meeting of 324 persons a motion was carried by a 
majority of 34, all voting. How many voted aye, and how many 
no? 

4. A father said to his son, " After 3 years I shall be 3 times as 
old as you will be, and 7 years ago, I was 7 times as old as you 
then were." What were the ages of father and son ? 

5. A farmer paid 4 men and 6 boys $10.80 for laboring one 
day, and afterwards he paid 3 men and 9 boys $12.15 for one 
day. What were the wages of a man and what the wages of a 
boy? 

6. If 1 is added to the numerator of a fraction, the resulting 
fraction will be equal to one fourth ; but if 1 is added to the 
denominator, the resulting fraction will be equal to one fifth. 
What is the numerator, and what is the denominator, of the 
fraction ? 



164 SIMULTANEOUS EQUATIONS 

7. What fraction is that whose numerator being doubled and 
denominator increased by 7, the value becomes f; but the de- 
nominator being doubled and the numerator increased by 2, 
the value becomes -| ? 

O 

8. Thirty feet more than ^ of the present height of the highest 
of the Egyptian pyramids is equal to 1 of its original height; 
300 ft. less than 6 times its present height is equal to 5 times its 
original height. Find the original and present height. 

9. Five dry quarts and 10 liquid quarts equal 15 liters, and 
7 dry quarts and 4 liquid quarts equal 11^ liters. How many 
liters are there in a dry quart and how many in a liquid quart ? 

10. One metric ton and 400 kilos equal 3080 lb., and 4 metric 
tons and 100 kilos equal 9020 lb. How many pounds in a kilo ? 

11. Fifty laborers were engaged to remove an obstruction on 
a railroad ; some of them by agreement were to receive $ 1.80 per 
day and others $3. There was paid them just $ 96. No memo- 
randum having been made, it is required to find how many 
worked at each rate. 

12. A merchant received from one customer $26 for 10 yd. of 
silk and 4 yd. of cloth ; and from another customer $ 23 for 
7 yd. of silk and 6 yd. of cloth at the same prices. What was 
the price of the silk and of the cloth ? 

13. A mechanic and an apprentice together receive $40. The 
mechanic works 7 days and the apprentice 12 days; and the 
mechanic earns in 3 days $7 more than the apprentice earns in 
5 days. What wages per day does each receive ? 

14. I have 7 silver balls equal in weight and 12 gold balls 
equal in weight. If I place 3 silver balls in one pan of a balance 
and 5 gold balls in the other, I must add to the gold balls 7 
ounces to maintain equilibrium. If I place in the pan 4 silver 
balls and in the other 7 gold balls, the balance is in equilibrium. 
What is the weight of each gold and of each silver ball ? 

15. When the greater of two numbers is divided by the less, 
the quotient is 4 and the remainder 3 ; and when the sum of the 



PROBLEMS TWO UNKNOWN NUMBERS 165 

two numbers is increased by 38 and the result divided by the 
greater of the two numbers, the quotient is 2 and the remainder 2. 
SUGGESTION. To find the quotient, subtract the remainder from the divi- 
dend and then divide by the divisor. 

16. Henry expended 95 ^ for apples and oranges, paying 5 f 
for each orange and 4 ^ for each apple. If he had 22 of both, 
how many of each did he buy ? 

17. The admission to an entertainment was 50 f for adults and 
35 ^ for children. If the proceeds from 100 tickets amounted to 
$ 39.50, how many tickets of each kind were sold ? 

18. An apple woman bought a lot of apples at 1 f each and a 
lot of pears at 2 ^ each, paying $ 1.70 for the whole : 11 of the 
apples and 7 of the pears were bad, but she sold the good apples 
at 2^ each and the good pears at 3 $ each, receiving $2.60. 
How many of each fruit did she buy ? 

19. A boy bought 570 oranges, some at 16 for 25^ and the 
remainder at 18 for 25 p. He sold them all at the rate of 15 for 
25 ^ and made a profit of 75 ^. How many oranges at each price 
did he buy ? 

20. Up to 1908 the highest speed for a short distance on a 
railroad was made March, 1901, by a train on the Plant System 
between Fleming and Jacksonville, Fla. The next highest 
was made July, 1904, on the Philadelphia and Reading, between 
Egg Harbor and Brigantine. The time being taken the same for 
both, the distance run by the first train was to the distance run 
by the second train as 25 is to 24, and the difference in rates in 
miles per hour was 4.8. What was the rate of each train in 
miles per hour? 

SUGGESTION. The time being the same, distances run are proportional to 
rates. 

21. A gasoline launch made a trip up a river and back in 4 hr. 
Had it made a trip 1 mi. more in total length on a lake, it would 
have been gone only 3| hr. The rate of the launch in still water 
was 3 times as great as the rate of the river current. What was 
the round trip distance, and what the rate of the launch ? 



166 PROBLEMS 

22. The sum of the two digits of a number is 8, and if 36 is 
added to the number, the order of the digits is reversed. Find 
the number. 

SUGGESTION. In order to represent numbers in the Arabic scale by letters, 
we must proceed as follows : 

Let t = the number represented by the figure in tens' place, 
and u -. - the number represented by the figure in units' place. 

Then 10 + u = the number, and 10 u + t the number when the figures 
are reversed. 

23. Find a number which is greater by 2 than 5 times the sum 
of its digits, and if 9 is added to it, the digits will be reversed. 

24. If a certain number is divided by the sum of its two digits, 
the quotient is 6 and the remainder 3; if the digits are interchanged 
and the resulting number is divided by the sum of the digits, the 
quotient is 4 and the remainder 9. What is the number ? 

25. Find that number of two digits to which if the number 
found by changing its digits is added, the sum is 121 ; and if one 
of the two is taken from the other, the remainder is 9. 

26. Two persons 27 miles apart, setting out at the same time, 
are together in 9 hr. if they walk in the same direction ; but if 
they walk in opposite directions towards each other they meet 

.in 3 hr. Find their rates. 

27. Two trains set out at the same moment, the one to go from 
Boston to Springfield, the other from Springfield to Boston. The 
distance between the two cities is 98 miles. They meet each other 
at the end of 1 hr. and 24 min., and the train from Boston travels 
as far in 4 hr. as the other does in 3 hr. What is the speed of 
each train ? 

28. A pound of tea and 10 Ib. of sugar cost $1. But if tea 
were to rise in value 10% and sugar 20%, they would cost $1.15. 
Find the price per pound of each. 

29. A boat's crew can row 15 mi. an hour downstream. The 
crew can row a certain distance in still water in 15 min. and 
require 20 min. to row the same distance up the stream. Find the 
rate of rowing in still water and the rate of the stream. 



PROBLEMS --TWO UNKNOWN NUMBERS 167 

* 30. If the sides of a rectangular field were each increased by 
2 yd., the area would be increased by 220 sq. yd. If the length 
were increased and the breadth diminished each by 5 yd., the 
area would be diminished by 185 sq. yd. What is the area? 

31. A rectangular field has the same area as another which 
is 6 rods longer and 2 rods narrower, and also the same area 
as a third which is 3 rods shorter and 2 rods wider. Find its 
dimensions. 

32. The circumference of the large wheel of a carriage is 55 in. 
more than that of the small wheel. The former makes as many 
revolutions in going 250 ft. as the latter makes in going 140 ft. 
Find the number of inches in the circumference of each wheel. 

33. A man invests $10,000, part at 41% and the rest at 31%. 
He finds that 6 yr. interest on the first investment exceeds 5 yr. 
interest on the second by $1098. How much does he invest at 
each rate? 



34. The report of an explo- o- = vekc. of .sound 

sion traveled 5560 ft. in 5 sec- - 
onds with the wind, and 3204 ft. y= Vl ' loc - of Avin<1 



in 3 seconds against the wind. ./ + //= vi-i...-. with win.i. 

Find velocity of sound in still > 

air and velocity of wind. *~ y= veloc ' affalnst wiml 

35. Two men can do a piece of work in 30 hr. ; they can also 
do it if the first man works 251 hr. and the second 324 hr. In 

o O 

how many hours can each alone do the work ? 

36. Portland cement, when pure, contains 65 % lime and 35 r / 
silica and aluminium, that is to say, 10 cu. ft. of it would contain 
6.5 cu. ft. lime and 3.5 cu. ft. silica and aluminium. If made 
from limestone rock containing 60% lime and 24% silica and 
aluminium and clay containing 2% lime and 95% silica and alu- 
minium, how many cubic feet of limestone and how many cubic 
feet of clay should be mixed and burnt together to make 10 cu. ft. 
of cement? Ans. 10.8 cu. ft. and .95 cu. ft. 



168 SIMULTANEOUS EQUATIONS 

SIMULTANEOUS EQUATIONS WITH THREE UNKNOWNS 

105. Simultaneous Equations containing Three or More Unknown 
Numbers. Such equations are solved by repeating the process of 
elimination. 

1. (1) x-2y + 3z = 6, 

(2) 2.4-3?/-4z 

(3) 3x-2y + 5z 

On inspection we perceive that of the three unknown numbers 
x can most readily be eliminated. 



(2) 2x+3y- 4z 

2 .c - 4 ?/ + 6 z = 12 (Mult. Ax.) 



(4) 7y-10z= 8 (Sub. Ax.) 

(3) 3x-2y+ 50 = 26 

(1 2 ) 3 x - 6 y + 9 z = 18 (Mult. Ax.) 

(5) 4?/- 4z = 8 (Sub. Ax.) 

y - z= 2 (Div. Ax.) 



Taking now equations (4) and (5j), we proceed as in solving 
simultaneous equations containing two unknowns. ( 100.) 

(5 2 ) 7 y - 7 z = 14 (Mult. Ax.) 
(4) 7y-10z= 8 

3 z = 6 (Sub. Ax.) 

z 2 Ans. (Div. Ax.) 

(5^ y 2 = 2 (Substituting its value, 2 ; for 2.) 

?/ = 4 ^4??s. (Sub. Ax.) 

(1) x - 2 X 4 + 3 x 2 == 6 (Substituting their values for 

a 8 + 6 = 6 y and z) 

x = S Ans. 



THREE UNKNOWN M .MI5EKS 



169 



VERIFICATION. Since the value of x was just found from (1), 
it is scarcely necessary to verify in (1). 

(2) 2xS+3x4-4x 2= 20, or 16 + 12 - 8 = 20, 

(3) 3 x a- 2 x 4 + 5 x 2 = 26, or 24 - 8 + 10 = 26. 
Observe tha.t we 

(1) Begin by choosing the unknown that can be most easily eliminated. 

(2) Eliminate this unknown between two of the equations, getting 

equation (4), and then eliminate the same unknown between 
another pair of the given equations, getting equation (5). 

(3) Take equations (4) and (5) and solve as in 100, getting the 

values of both unknowns. 

(4) /Substitute the two answers found in one of the given equations, 

getting the value of the unknown first eliminated. 

(5) Verify in the other two equations. 



2. 



4. 



4z=29, 



3. 



x + 3y + 2z = 25. 
x + y + z = 6, 



3x-2y + z = 2, 



5. 



x 



3y + 3z = 25. 

2y z-- 6, 



6. 



-42 = 98, 
9 x 15 y + 6 z = - - 6, 
80? 3 9=12. 



7. 



5 x - - 11 ?/ z = 7, 



8. 



9. 



-- 



= -20, 



-8 =-27. 



106. Special Processes of Elimination. 

Yl) 3 x + 4 11 = 25 SI-GGESTION. Combine (1) and (2), 

1 (2} 5 ?/ -I- f>"z = 50 eliminating y, and getting equation (4). 

Then taking (-4) and (3) together, solve 

J 



107. Review. 1. Sim. Eq's.; 2. Elimination; 3. Methods of Eliin. 



170 PROBLEMS IN THREE UNKNOWNS 

108. Problems involving Three Unknowns. 

1. Determine three numbers such that their sum is 9 ; the sum 
of the first, twice the second, and three times the third, 22; and the 
sum of the first, four times the second, and nine times the third, 58. 

2. A and B together possess only -| as much money as C ; B and 
C together have 6 times as much as A ; and B has $ 680 less than 
A and C together. How much has each ? 

3. Three men bought grain at the same prices. A paid $5.70 
for 2 bu. of rye, 3 bu. of wheat, and 4 bu. of oats ; B paid $ 7.40 for 
3 bu. of rye, 5 bu. of wheat, and 2 bu. of oats ; and C paid $ 6.20 
for 2 bu. of rye, 4 bu. of wheat, and 3 bu. of oats. What was the 
price of each ? 

4. Divide 800 into three parts such that the sum of the first, 
of the second, and -| of the third shall be 400 ; and the sum of 
the second, f of the first, and ^ of the third shall be 400. 

5. A quantity of water sufficient to fill three jars of different 
sizes, will fill the smallest jar four times ; the largest jar twice 
with 4 gal. to spare; or the second jar three times with 8 gal. to 
spare. What is the capacity of each jar in gallons ? 

6. The average age of three persons is 41 yr. The average 
age of the first and second is 37 yr., and of the second and third 
is 29 yr. Find their ages. (See 153.) 

7. A number is expressed by three figures whose sum is 11. 
The figure in the place of units is double that in hundreds' place, 
and when 297 is added to this number the sum obtained is ex- 
pressed by the figures of the number reversed. What is the 
number ? 

SUGGESTION. Let x = figure in hundreds' place, y = that in tens' place, 
z = that in units' place. Then the number is 100 x + 10 y + z. (See 104, 
22.) 

8. The middle digit of a number of three figures is one half 
the sum of the other two digits. If the number is divided by 
the sum of its digits, the quotient is 20 and the remainder 9 ; 



PROBLEMS 171 

and if 594 be added to the number, the digits will be reversed. 
Find the number. 

9. In round numbers in 1904 New York City, Boston, and 
Chicago paid out to the public together 20.4 millions of dollars 
merely as interest on the indebtedness of these cities. Three 
times Boston's amount increased by twice Chicago's equals 2 
millions more than New York's ; and 3 times Boston's and 
Chicago's together is 4.4 millions more than New York's. What 
was the interest paid by each in round numbers ? 

10. For running expenses in 1904 the cities of Washington, 
D.C., Minneapolis, Minn., and Indianapolis, Ind., paid together in 
round numbers 12 million dollars. Twice Indianapolis's amount 
increased by Minneapolis's was $800,000 more than Washing- 
ton's, and the Washington and Minneapolis sums together were 
$ 1,200,000 less than 5 times Indianapolis's. 

What was the running expense of each? 

11. If a circle is inscribed in a triangle 
as shown in the figure, it is proved in 

geometry that /( \B 

AF = AD, DB = BE, CE = CF. 

Now if AB = 12 in., EC = 8 
in., and AC 16 in., find the 
lengths of AD, BE, and CF. 

12. A circle is inscribed in a triangle ABC, as in the figure, 
whose perimeter (distance around it, or AB -f- BC -f- CA) equals 
42 ft. Now AF exceeds DB by 6 ft., and CF exceeds DB by 3 ft. 
Find the sides of the triangle. 

13. The total value of farm property of continental United 
States according to the census of 1900 was in round numbers 20.5 
billions of dollars. The land exceeded double the value of the 
buildings added to the value of the live stock and implements by 2 
billions ; and the land and buildings together exceeded 4 times the 
live stock and implements by ^ billion. What was the value of 
the land, buildings, and live stock and implements respectively ? 




CHAPTER VI 

uRAPHS AND GRAPHICAL SOLUTION OF SIMPLE EQUATIONS 

109. Graph Paper is paper ruled very accurately into small 
squares or parallelograms. (See graph paper ruling in the dia- 
gram of the next article.) Formerly such paper was very ex- 
pensive, but of late years it has been manufactured in increasingly 
large quantities, and is now sold very much cheaper. 

The paper is commonly ruled in units of the metric system, the 
centimeter being the standard unit. A centimeter is one hun- 
dredth of a meter, just as a cent is one hundredth of a dollar. 
A millimeter is one tenth of a centimeter or one thousandth of a 
meter, just as a mill is one thousandth of a dollar. Now a meter 
is somewhat longer than a yard, since it equals nearly 39.37 inches. 
Hence a centimeter is .3937 inch, or very close to A inch. A milli- 
meter is one tenth of A inch, or .04 inch, or -^ of an inch. 

The larger squares on the graph paper are square centimeters, 
and the little squares are therefore 2-millimeter squares. 

110. Location of Places by Reference to Two Standard Lines. As 

an example of such location of places take the case of townships, 
or towns, as they are often called. 

In the figure, p. 173, let the centimeter squares denote town- 
ships, XX 1 being a "base line," and TY' a " principal meridian." 
Then any township, as the one marked with the letter A, is 
described as Town 3 North, Range 2 East; the township marked 
B, as Town 4 North, Range 3 West ; the township marked (7, as 
Town 2 South, Range 1 West; etc. Townships are divided, of 
course, into 36 sq. mi., and not into 25 squares, as in figure. 

172 



AXES 



173 



o 



A 



Y- 



We desire here, however, to locate points, and not squares. In 
the West, roads usually run on section lines. If a village is 
at a cross-roads, any 
other cross-roads can 
be located by saying 
it is so many miles 
east or west, and so 
many miles north or 
south of the village. 
Similarly in a city 
in which the streets 
run east and west, 
and north and south, 
a street corner can 
be located in the 
same way by refer- 
ence to some square. 

As another exam- 
ple, the location of 
places on the earth's 
surface can be taken. 
Every such point 
can be located by 
saying it is such and such a number of degrees north or south of 
the equator, and such and such a number of degrees east or west 
of the standard meridian. 

111. Axes. Two lines are taken at right angles as lines of ref- 
erence, and are called axes. The horizontal one, XX' (see figure), 
is called the X axis, and the vertical one, FF', the F axis. Dis- 
tances measured to the right of FF' are positive ; those to the 
left of FF' are negative. Distances above XX' are positive; 
those below XX ' are negative. The point where the axes cross 
is called the origin. 

The axes divide the diagram into four parts called quadrants. 
A is in 1st quadrant, B in the 2d, C in 3d, and F in 4th. 



174 



GRAPHS 



112. Location of Points. Any point in the plane of XOY, as P 1 
(see figure below), is located by measuring the perpendiculars, or 
shortest distances, from this point to the reference axes. Dis- 
tances measured parallel to the X axis, as 2\B, are denoted by x, 



-X 



Unit 



-IV*- 



Y- 



Pr 



fl 



-A- 



p; 



and are called abscissas; and distances parallel to the Faxis, as 
PiAj are denoted by y, and are called ordinates. The abscissa 
and ordinate of any point are called the coordinates of the point. 
Centimeter being unit of measure, the coordinates 

of PI are 2, 1 (x = 2, y = 1) ; that is PB = *2, PA = 1 ; 
of P 2 are - 2, .4 (x - - 2, y = .4) ; that is PoO = - 2, P 2 D = A ; 
of P 3 are - - 1.6, --\A (x= -- 1.6, y = - 1.4) ; for signs see 111 ; 
of P 4 are 3, -- 2.8 (x = 3, y = - 2.8) ; for signs see 111 ; 
of P 5 are 0, - 2 (x = 0, y = 2) j for sign see 111. 



LOCATION OF POINTS 175 

1. Write the coordinates of the points located by the letters 
a, b, c } d, e, /, g, h on the diagram. 

2. Take a piece of squared paper and draw two heavy lines 
(on two centimeter division lines) to represent the two axes, 
just as was done in the two diagrams already given. Now locate 
the following points by a dot on the diagram, putting a small 
letter beside the point located, writing a beside the first point, b 
beside the second, and so on. 

a. (x=2,y=4:). * b. (# = 4, ?/ = 4). c. (x=-3, 2/=3.4). 

d. 0=7, y= --1). e. (x = 3, y = 4). /. (x = 2.2, y = 0). 
g. (#= 1.8, #= .6). h. (x = Q, 2/ = 2.5). (a; = 2, y = 0). 

3. Make rule for locating points, their coordinates being given. 

(1) Where does one start from to count out the units in x? 

Ans. From the origin at 0. 

(2) If x is given positive, in which direction is it measured ? If 

x is negative, in which direction is it measured ? 

(3) Having arrived at the extremity of x, from there in which 

direction is y measured if it is positive ? If negative ? 

(4) Where is the point sought now found ? 

4. Make another diagram by drawing axes at another place 
on the sheet of squared paper, and locate the following points, 
understanding that the first number inside any parenthesis gives 
the value of the abscissa, or x, and the second number the value 
of the ordinate, or y. 

a. (1, 1). 6. (2, 4). c. (- 2, 1). d. ( - 3, - 3). 

e. (-3, 6). /. (- 2, - 4). g. (- 5, 0). h. (7, - 2). 
i. (0,0). j, (0, -3). k. (-2, -8). 1. (10,11). 

113. Graphs. If a series of points is " plotted " (that is, 
located and marked on a diagram) representing values of a quan- 
tity that changes, and these points are joined by a running line, 
this line is called the graph of the law or data that determined the 
points. (See 73. It can replace 113, if teacher so desires.) 



176 



GRAPHS 



1. Graphs can be used to show to the eye changes in tempera- 
ture. The following graph is obtained from the data contained 
in the table at its right. The hours counting from 12 noon are 
abscissas, and the number of degrees at these hours are the ordi- 
nates of the points marked on the figure. 



in- 



-12J 



LE..M, 



r > I. \ 



\ 



M, 



12 M. 


15 


3 P.M. 


17 


6 P.M. 


14 


9 P.M. 


11 


12 A.M. 


9 


3 A.M. 


6 


6 A.M. 


4 


9 A.M. 


9 


12 M. 


11 



2. Construct a similar graph being given the following data : 
12 M., 28; 3 P.M., 20; 6 P.M., 14; 9 P.M., 9; 12 P.M., 7;3 A.M., 
5; 6 A.M., 3; 9 A.M., 11; 12 M., 18. 

3. Construct the graph for the following readings actually 
taken at a certain place in the month of December : 12 M., 28 ; 
3 P.M., 10; 6 P.M., 4; 9 P.M., -1; 12 P.M., -5; 3 A.M., -6; 
6 A.M., -7; 9 A.M., -3; 12 M., 5. 

4. The graph, p. 185, made from the data below, shows changes 
in Public Debt of the United States from 1830 to 1907. 

1830, $49 millions. 1864, 1816 millions. 1894, 1632 millions. 

1840, 4 millions. I860, 2773 millions. 1900, 2136 millions. 

1850, 63 millions. 1870, 2481 millions. 1907, 2492 millions. 

1860, 65 millions. 1880, 2120 millions. 

1862, 524 millions. 1890, 1552 millions. 



GRAPHS OF STATISTICS 



177 



In this problem it is convenient to let a centimeter along the X 
axis represent 10 years, and a centimeter along the Y axis repre- 
sent 500 million dollars. 



Mill ons 



\ 



200(1 



\ 



1500 



1000 



X- 



Y' 



1830 



J840 



1850 



1860 



1870 



1880 



1890 



1900 



This graph tells the story of the Civil War in a most vivid way. 

5. Construct the graph corresponding to the following table of 
figures, representing expenditures in round numbers of the United 
States government for every fifth year, 1870-1905. 

1870, $294,000,000. 1885, .$260,000,000. 1900, $488,000,000. 
1875, 275,000,000. 1890, 298,000,000. 1905, 567,000,000. 
1880, 265,000,000. . 1895, 356,000,000. 

6. Construct graph for Post-office Expenditures, 1865-1905. 

1865, $15,000,000. 1880, $37,000,000. 1895, $ 90,000,000. 

1870, 23,000,000. 1885, 49,000,000. 1900, 110,000,000. 

1875, 36,000,000. 1890. 09,000,000. 1905, 169,000,000. 

SUGGESTION, Put one T centimeter = $20,000,000. 



178 



GRAPHS 



7. Compare by the graphical method the growth of population 
of the United States and the United Kingdom (England, Ireland, 
and Scotland), 1800-1900. 

STATISTICS OF POPULATION OF UNITED STATES IN ROUND NUMBERS 

1800, 5,300,000 1840, 17,100,000 1880, 50,200,000 

1810, 7,200,000 1850, 23,200,000 1890, 62,600,000 

1820, 9,600,000 1860, 31,400,000 1900, 76,300,000 

1830, 12,900,000 1870, 38,600,000 

STATISTICS OF POPULATION OF UNITED KINGDOM IN ROUND NUMBERS 

1800, 15,700,000 1841, 27,000,000 1881, 35,000,000 

1810, 17,600,000 1851, 27,000,000 1891, 38,000,000 

1820, 21,000,000 1861, 29,000,000 1901, 42,000,000 

1831, 24,000,000 1871, 31,000,000 

Put both graphs on the same figure, using the same axes for both. 

8. Construct a graph for the movement of train No. 25 on the 
Boston and Maine which ran on. the time card annexed. 



hr 



hi, 



2 



Miles 



10 



20 



30 



40 



50 



MILES STATIONS 

Boston 

4 
10 
20 



Cambridge 

Waltham 

Concord 



36 Ayer 

50 Fitchburg 



A.M. 

4.40 
4.49 
5.00 
5.17 
5.58 
6.37 



Where the graph consumes time faster than distance, the train 
is going slower ; where the reverse is true, it is going faster. 

9. A train on the Pennsylvania Railroad ran on the following 
schedule : Pittsburg, 8 A.M. ; Irwin, 21 mi., 8.43; Latrobe, 40 mi., 




RENE DESCARTES (1596-1650) 



GRAPHS OF SIMPLE EQUATIONS 179 

9.18 ; Johnstown, 76 mi., 10.17 ; Cresson, 99 mi., 11.01 ; Altoona, 
114 ini., 11.36; Huntingdon, 148 mi., 12.34 P.M.; Newport, 218 
mi., 2.16 ; Harrisburg, 245 mi., 3.00. Construct a graph for its 
movement, letting 1 cm. = 25 ini. and 1 cm. = 1 hr. How do you 
account for its irregularities ? 

10. Scientists construct all sorts of apparatus to register graphs 
automatically. A good illustration is the thermograph (see 
drawings in Century and Standard Dictionaries). A cylinder 
covered with paper is revolved by clockwork, and a pen rules a 
line on it as it revolves. The position of the pen is controlled by 
the temperature of the air. As a piece of metal to which the pen 
is attached expands or contracts, it moves the pen up or down. 
The graph is a continuous wavy line, the high parts of which 
show a higher temperature and the low parts a lower temperature. 
The hours of the day are printed on the paper. 

Other graph apparatus that may be mentioned are the baro- 
graph, the seismograph, the anemograph, etc. 

11. Graphs are used by writers on history and economics to 
show changes in population, expenditures, production, etc. ; by 
physicists, scientists generally, engineers, etc., to express laws of 
nature, the working of machinery, etc. ; by business houses to show 
changes in prices, sales, cost of production, etc. In short, graphs 
have a wide range of uses, and the student should learn well how 
to construct and read them. 

114. Graphs of Simple Equations containing Two Unknowns. 
An equation containing two unknowns, as 3 x -+ 4 y 9, can have 
a graph, because for every value of x the equation gives a cor- 
responding value of ?/, and for every such pair of values we get a 
point located ; joining several such points, a portion of a graph 
results, called the graph of the equation. The graph of an equa- 
tion has been called its geometric picture. 

The idea of constructing graphs of equations was conceived by 
the French philosopher Descartes and published to the world in 
the year 1637 A.D. The full development of this idea is called 



180 



GRAPHS 



analytic geometry, which is a college study. In analytic geome- 
try a graph is called the locus of a point. 

1. Construct the graph corresponding to the equation 

2 x + 3 y = 6. 

SOLUTION. We first solve the equation for y in terms of x : 

y = 6 ~ 2x (Axioms ?) 



We next find values of y which correspond to assigned values of x in this 
way: 

Let 



x 4 ; then y=- - .67 (to nearest hundredth). 

o o 



(\ 9 v 3 

x = 3 ; then y = - = 0. 

o 

In the same way we get the value of y corresponding to other assigned 
values of x. The results of all these calculations are put in the following 
table, and the points located by the sets of coordinates in the table are 
marked on the graph diagram, and a line is drawn through them. Each 
decimal value is taken to the nearest hundredth. 

TABLE 



s 






























Y 








































// 


"S 


x 








































































X 








































































^ 


x 




































































y 




x 








































































s 


s 








































































X 






































































I 


^ 


~^ 








































































S 














































II 


























s 


*N 










































\ 




























(' 


\ 










































\ 
































s 








































\ 
































X 


s 






































\ 
































? 


*s 






































\ 






























/ 




X 






































\ 


































V 


\ 




































\ 




































s 




































\ 


































(' 


s 


x 


































\ 






































X 
















/> 


















\ 






































V, 


\ 










K 


















/ 


I 


\ 








o 






























', 


X 
































\ 








































s 


s 






























\ 










































X 






























\ 










































5 






























\ 






































































7 


k, 




































































A 




N 








































































\ 








































































\ 
































































i 








\ 

































( *, y) 


PT. 


( 4, 


-.07) 


a 


( 3, 


) 


b 


( 2, 


.67) 


c 


( 1, 


1.33) 


d 


( > 


2 ) 


e 


(-1, 


2.67) 


f 


(-2, 


3.33) 


9 


(-3, 


4 ) 


h 



GRAPHS OF SIMPLE EQUATIONS 



181 



a. Notice that all the points lie in a straight lino, or would do so if the 
values of their coordinates were found to more decimal places. It w r ill be 
found that whenever the given equation is of the first degree (see 57), all 
the points lie in a straight line. The student should test this statement until 
he is satisfied of its truth. For this reason equations of the first degree are 
often called linear equations. 

Since two points determine a straight line, it will be necessary to find the 
coordinates of only two points, and not of more, as was done in Ex. 1, p. 180. 
As a rule the two points where the line crosses the axes (e and 6 in the pre- 
ceding figure) are found. This is done by letting x= and finding the cor- 
responding value of y, and then letting y = and finding the corresponding 
value of x. 

To check the result, locate a third point at the extreme edge of the diagram, 
and see whether the line drawn through the first two points goes through it. 

The student should each time make a table, writing its equation above it. 

2. Construct the graph of the equation x + y = 1. 
y =-x- 1 



(, >/) 


PT. 


(0, - 


1) 


k 


(-1, 


0) 


m 


(-3, 


2) 


n 



SOLUTION. The values are assigned as suggested in 
paragraph just given. The graph is constructed on 
the diagram to Ex. 1 of this article. Notice that the 
three points k, m, n lie in a straight line. 



Construct the graphs of the following equations, following Ex. 2 
as a model. 



3. 

7. = 



4. 



4. 

8. 



5. x 
9. = 



= 2. 6. 2x 3y = 6. 
. 10. 4 a;- 



11. Make a rule for constructing the graph of 'a simple equation. 

(1) What value is assigned to x to get corresponding value of y? 

(2) What value is next assigned to y to get corresponding value 

of ? 

(3) What is now done with the pairs of values found ? 

(4) After the two points are located, what line is drawn ? 

(5) How is the work checked ? 



182 



GRAPHS 



b. When the two points located by putting x and y in turn equal to zero 
are close together, they do not give a good determination of the graph. 
In this case get the check point first and draw the graph through it and the 
more remote of the other two, using the intermediate point to check the result. 



12. 
15. 



= 2x 9. 



13. 2 a; -3 # = 



16. 



14. 4 a; + 5 #=24. 



17. a; + # = 
20. 12a?-15# = 



= 2 x. SUGGESTION. Here two other 
points besides (0, 0) are needed. 

18. # = --4 --7 a. 19. 05 = 5 #+3. 

21. # = 4. SUGGESTION. Locate any two 
points whose ordinate is 4. The graph is a 
line parallel to axis of X. 

-3. 23. x = y. 24. 2a?4-#=20. 25. lla?4-13#= 6. 

115. Solution of Simultaneous Equations containing Two Un- 
knowns by Means of Graphs. This is accomplished by construct- 
ing the graph of each of the two given equations on the same dia- 
gram. The coordinates of the point of intersection of the two graphs 
are the answers sought, that is, are the values of x and # which 
satisfy both equations ( 98). 



22. 



1. Solve 



(1) 3 x -f 4 it = 8 
^ ' 

(A) 4:X u # = 5 



, 

by the graphic method. 

(1) 3aj + 



-x'- 







;? 



X 



N 



-x 



(0,2) 


a 


(2.7, 0) 


6 


(4, - 1) 


c 



(2) 4aj~6# = 5 



(0, -8) 


d 


(1.2, 0) 


e 


(4, 1.8) 


f 



SIMULTANEOUS EQUATIONS SOLVED BY GUAI'IIS 183 



The coordinates of the point of intersection, P, of the two graphs are 
x = 2, y = |, which are the values of x and y sought. 

To check the result, we solve the two given equations in the usual way. 
(See 100.) 

(li) 12 x + 16 y = 32 (Ax.?) (l)3x + 4x| = 8. 

(2i) 12 a- 18 y == 15 (Ax.?) .'. x = 2. (Check.) 

34 y=-. 17 (Ax. ?) 

y = \. (Ax. ?) 

Solve the following by the graphical method and check by one of 
the algebraic methods ( 99, 100). (If this chapter is studied 
before Chapter V, this way of checking, of course, cannot be used.) 

2. { 2 *+32/=12, 3 



(2x-y=3. 



5. 



8. \ 



llx-2y=2l, 
20+4 y=-18. 



1 3 a; 5 #=19. 



n 



6. 



9. 



12. 



(3x-2y=S. 

7. j X+ ' 
(x v 



3. 



f4aj-5y=10, 
l2<c+32/=9. 



05=3. 



10. 



13. 



14 J8 X 21 ?/ = 33, SUGGESTION. To put y =0 in (2) runs the graph 

I A /v_i_^^ i -1 77 off the sheet of paper. Put w = 4 instead. 
L o x -\- oj 2/ J- 1 ' 

REMARK. For such an equation as x + y = 1000, e.g., scale of 1 cm. = 100 
is needed. 



15. 



x _y = 
2 3" 

x_ 
13 "^ 



16. 



2 3 



17. 



a; 



6 "2 

3_L/ = 
10 " 



U 3 

18. Construct on same diagram graphs of the equations, 
2 x + 3 y = 4, 2x+y=12,x + 3y=--4:, 

REMARK. Because all three equations are satisfied by the same values of 
x and ?/, their graphs meeting in the same point, they are said to be consis- 
tent or compatible equations. 



184 GRAPHS 

*19. Solve 7 x + 2 y = 53, 12 a + 5 y = 94, 14 & - 11 y = 76 

graphically. 

20. Construct the graphs of (1) 2 # + y=4, (2) aj + 2 y = 4, and 
(3) 3 a- 5^ = 15. 

REMARK. These graphs meet in three points. They therefore do not 

constitute a consistent or compatible system of simultaneous equations. 

What values of x and y satisfy (1) and (2) ? What values satisfy (1) 
and (3) ? What values satisfy (2) and (3) ? 

21. Construct the graphs of 3 x 5 y = 15, and 6 x 10 y = 60. 

REMARK. These graphs are parallel and will not meet. Hence there is 
no solution. In attempting to solve the equations by the algebraic method both 
x and y are eliminated at the same time, the resulting statement being = 30. 
These two equations are inconsistent or incompatible. 

22. Construct the graphs of 4 x 7 y = 6, and 12 x 21 y = 18. 

REMARK. These graphs coincide, and every pair of values that satisfies 
one of the equations satisfies the other. The equations are in fact the same 
equation, the second form being obtained from the first. How ? 

This problem throws light on the marking of equations used in this book. 
See 99, a. 

23. If a third equation is obtained from two given ones, as 
was done in 103, will the graph of the new equation pass 
through the intersection of the first two? 

24. Suppose (1) 5 x -\- 3 y = 10, (2) 8 x 5 y = 16 are two given 
equations and (3) is obtained from (1) and (2) by adding them. 
Also that (4) is obtained by subtracting (1) multiplied through 
by 3 from (2) multiplied through by 2. Will (3) and (4) inter- 
sect at the same point as (1) and (2)? 

25. Without constructing them can you tell whether the 
graphs of 6 x 4 y = 12 and 3x Q = 2y cross or are parallel 
or are the same ? How do you know ? 

26. What can you say of the graphs of ox=ly-\-12 and 



CHAPTER VII 

SIMPLE QUADRATIC EQUATIONS. LITERAL EQUATIONS. 
GENERALIZATION. INDETERMINATE EQUATIONS. NEG- 
ATIVE SOLUTIONS. --HARDER FACTORING 

116. The Trinomial Square Formula. The formula 

(a + b) 2 = a 2 + 2ab + b 2 

is used many times, and in numerous ways. At first we had 
(tt -f- b) 2 given to find the right member ( 45). Then in factoring 
trinomial squares we had the right member given to find the left 
member ( 52, II). Later ( 140) this formula will be used to 
make the rule for extracting the square root of polynomials. We 
expect to use it in two other ways in this chapter. 

1. Given a 2 + 2 ab to find b 2 . 

SOLUTION. We first extract the square root of the first term, a 2 , getting 
a ; next we double this root, getting 2 a ; then we divide the second term, 
2 a&, by 2 a, obtaining b ; last of all we square this quotient, and thus 
obtain 5 2 . 

Find the third term in each of the following : 

2. a 2 -2ab. 3. a 2 + 6a. 4. a 2 --10a. 
5. y* + l2y. 6. x*-2x. 7. x 2 -2cfx. 
8. ar + 18a. 9. x 2 + 4 x. 10. 

11. 25 z 2 - 30 a. 12. 49 x 2 - 28 xy. 13. 

14. 4a 2 -f-4a;. 15. Qx 2 -6x. 16. l--8xr. 

17. x 2 4- x. 18. x 2 + x. 19. a 2 b 2 - f ab. 

20. Given a 2 + b 2 to find 2 ab. 

SOLUTION. We extract the square root of the first term, a 2 , getting a ; 
next we extract the square root of the last term, 6 2 , getting b ; then we take 
twice the product of these two roots, getting 2 ab, 

185 



186 SIMPLE QUADRATIC EQUATIONS 

Find the middle term in the following: 

21. 4 a 2 + 9 b 2 . 22. a 2 + 9 b 2 . 23. 4 z 4 + 25 #*. 

24. x 4 + 64?/ 4 . 25. 25 a 4 + 9 b* 26. 36m 4 + 49n 4 . 

27. a 4 + l. 28. a 4 6 4 + T L. 29. 8 + 6 8 . 

30. a 6 + 6 6 . 31. 9ar + l. 32. 



SIMPLE QUADRATIC EQUATIONS 

117. Simple Quadratic Equations. Quadratic equations were 
denned in 57, and were solved in that article by factoring. We 
proceed to solve them here in a different way. 



SOLUTION. 7 x 2 3 x 2 = 88 + 12. (Sub. Axiom ; unknowns to left 

member, knowns to right.) 
4 x 2 = 100. (Collecting terms.) 

se a = 25. (Ax. ?) 

x = + 5. (Root Ax., 83, 6.) 

VERIFICATION. 7(5) 2 - 88 = 3 (5) 2 + 12 ; also 7(-5) 2 - 88 = 3(-5) 2 + 12. 

2. 3z 2 -2 = 2^ + 2. 3. ^-3 = 4z 2 - 



4. or 2 -36 =-+12. 5. ar* - 3 = 4a;2+18 . 

4 9 

6. (x + I) 2 = 2 x + 17. 7. (10 + a?)(10 - x) = 19. 

33 150 3 

8. -+- - =8. 9. 4 05 -=x -- . 

1 + x 1 x x x 

10. (a; + I) 3 - (a? I) 3 = 26. 11. (3 x + 2) x - 7 = (x + I) 2 . 
12. x- + 4 x = 5. 

SOLUTION. The solution is accomplished by a method called " completing 
the square." We consider x 2 + 4x as the first two terms of a trinomial 
square ( 116) corresponding to a 2 + 2 ab in a' 2 + 2 ab + b' 2 . To find 6 2 , we 
extracted the square root of a 2 , and divided 2 ab by twice this root, getting b. 
Then 6 was squared and added to a 2 + 2 ab. We proceed in the same way 
with x 2 + 4 x. 

= a; ; 4 a; -*- 2 a; = 2; 2 2 = 4. 

(Add. Ax., 4 being added to each side 

+ 4 = : 4 + 5, or 9. of the given equation.) 



SOLUTION OF QUADRATIC EQUATIONS 187 

(Root Ax. by extracting the square 

root of the two equals.) 
x = 2 -+ 3, or 2 3. (Sub. Ax., transposing 2.) 

x 1, or 5. Ans. 

VERIFICATION. (I) 2 + 4(1)^5 ; also (-5) 2 + 4(-5) =5. 

Solve and verify both answers in the following : 

13. 0* + 10 a; = 11. 14. x 2 - 14 x = 32. 

15. x 2 + 20x = -19. 16. x 2 -6x = 27. 

17. a 2 8a + 15 = 0. 18. x*--4:X = 77. 

19. 4z 2 + 12a+8 = 0. 20. 10 x- + 24 x = 27. 

21. 25x 2 + 20a = 12. 22. 9 # 2 + 24 a; =-15. 

23. Make a rule for solving quadratic equations of the kind 
given after Ex. 12. (1) What is added to each side of the equa- 
tion ? Ax.? (2) What is then done? Ax.? (3) What is the 
last step? Ax.? (4) Verification. 

24. aj* - 14 a; + 45 = 0. 25. a 2 - 12 aj + 35 = 0. 
26. 4 a 2 + 4 = 224. 27. 81 a 2 - 90 a; = 231. 

28. Solve problems 2-4, 13-27, by the factoring method, 57. 

29. Divide the number 70 into two such parts that their 
product may be 864. Prove. 

30. The square of a number increased by 14 times the number 
equals 275. Find the number. Prove. 

31. If you add 10 ft. and 6 ft. respectively to the dimensions of 
a certain square room, the resulting area is 3 sq. ft. more than twice 
the original area. How many feet in a side of the square ? 

32. What number is 4 times its reciprocal ? 

33. A rectangular field contains 240 sq. rd., one side being 
8 rd. shorter than the other. Find its dimensions. 

34. Two men, A and B, together can do a piece of work in 2 
days. If B requires 3 more days to do it in than A, how long 
will it take each of them ? 



188 LITERAL EQUATIONS WITH ONE UNKNOWN 

LITERAL EQUATIONS 

118. Literal Equations in One Unknown. A literal equation is 

one in which known quantities are represented by letters. 

The verification of a literal equation, when it leads to an identity 
( 76), shows that both the solution and the additions, multipli- 
cations, etc., in the verification have all been performed correctly. 

In the following exercise the student should both solve and 
verify. 

1 . ax + b 2 = a 2 + bx. 

(Sub. Ax. Unknowns to left mem 
SOLUTION, ax bx = a 2 b 2 . 

ber, and knowns to right.) 

(a - b)x = a 2 - V 2 . (Factoring, 50.) 

a = 1_|?. (Div. Ax.) 

a- b 

x = a + b. (Reduction to lowest terms, 60.) 

VERIFICATION. a(a + 6) + b' 2 = 2 + b(a + &). 

a 2 + ab + b 2 = a 2 + ab + 6 2 . 

2. ax-{-lr = bx+ab. 3. a-f-# b x. 

4. a x = b 8. 5. 4 (a? a) = 3x-5b. 

6. 2a-2c = 2b-2x. 7. 3 x-2(a -a) = 4 x-c. 

8. ax b = cx + d. 

SOLUTION, ax ex = b + d. (Sub. Ax., putting unknowns on left side 

and knowns on right side.) 
(a c)x = b + d. (On factoring in left member.) 



x = . (Div. Ax.) 

a c 



VERIFICATION, a (^} - b = e(^f\ + d. ( On substitutin g its value 

\a--cj \a-cj f or x in the given equation.) 

ab + ad , be + cd , ,*** I A . i 

b = - - + d. (Multiplying. ) 

a c a c 

ab + ad b(a c) = be + rd + d(a c). (Mult. Ax.) 
ab + ad ab + be = 6c + cd + ad dc. 
ad + bc = ad + ftc. 

9/v /> I /i /Y* /~i /^ 1 /^ 7^ o* I O /v o* n ^ /v '> ^ / 11 

tt*/ n C/// CiO J. V/ C/4*/ "t* . tt 1 *// tl/ O tc/U/ O A X 



SOLUTION OF LITERAL EQUATIONS 189 

12. a -l=' J -9. 13. !+* = . 14. * + * = <;. 

xx 1 -- x b a b 

a b a-\-b 3 ac bc 

15. - = - Ans. x = - 

x c x + 2c 26 



VERIFICATION. 



a ~ h = a + b (Substituting its value 

3 ac - be Sac -be for x m lven et l ua - 

2b 2b tion -) 

a - b a -f b 



3 ac be 2 be Sac be + 4 be 

. ) 



(Adding in the denomi- 



26 2b 

2 b(a -- 6) _ 2 b(a + 6) (Reducing the complex frac- 

3 c ( -- 6) 3 c (a + 6) tions to simple ones.) 

6bc=6bc. (00, and Mult. Ax.) 

1 4- 2 a a; + a 1 x -f- a ft # 5 a 

16. - -- = - 17. = 18. - ~+T* 

oxo xom b a a b 



J.J7 


& + 


~ ' 


L. <OU. 

a- 


x a b 


/ij.. 
X 


X 


22 


X 


b. 


23 Wia? 


-\-n a 


24. ^_1=^_ 


3 ab. 


xOO. 






a 




mo; 


n b 


a c 




25 


a 


b. 


26. - 


b 


fo; t? a 

/i / . ~ 


ex 




x 




05 


x 


a c b 


d 


2ft 


a i 


b 


oq a 


c 


ft* 

QA nr-i-h 4- 


1 






x 


fj\j 

C 


y a-y 


a 


& 




31. X 


_4_ x _l_ 


X (1 


35 


a x b xc- 


-X 



a b c a b c 

33. (a + 6)a; = m ~ ca?. 34. (a+a;)(6+a;)=(m4-a;)(?iH-a;), 



35. *- -=^ -. 36. 

p r q s cd C+ 

f rr\ / r\ 

37. a &-f -) = m(c--]. 38. 

V c \ bJ 



2 CX 



- . -- 

--- [- ^. 4U. 

bx ax a b a + b a 2 



ax 



190 LITERAL EQUATIONS 

NOTE. Equations from physics solved like those in the preceding article 
will be found on page 241. Familiarity with them is strongly advised. Their 
study should be taken up here, if there is danger that otherwise they might 
not be solved. 

119. Literal Equations containing Two Unknown Quantities. 

1. Given (1) x + y = s, (2) x y - d, to find x and y. 
SOLUTION, x -f y = s x + y =s 



2x = s + d (Ax.?) 2y=s-d (Ax.?) 

. = ii r=if ? 

, T x-, x s -f d , s d , n \ s + d s d __ -, 

VERIFICATION. (1) -^ - H -- - =s ; (2) --- - = a. 

22 22 

2. (1) x + y = 5a 4 5; (2) x y = 4a 5 &. 



3 x I 4 y ct 
3. SUGGESTION. Solve as in 100. Verify. 



4. \ 5. \ 6. -^ 

3 ?/ = c. [3o;+ wiy = q. 



f ax + 2 6w = 6. 

7. 8. (1) oo? + 6y = c ; (2) a'x + 6'y = c'. 

1 2 ax + 5 by = 9. See so]ution below> 

REMARK. It is often convenient, instead of using new letters, to use the 
same letters as before, but to mark them in some way. In this problem, 
instead of using three new letters, d, e, /, we have written in their place a', 
6', c', read ' a prime,' ' b prime,' ' c prime.' The a' means a number entirely 
different from a. The marks a", a'", etc., are also used, read, ' a second,' 
' a third," 1 etc. Also ai, ao, s are used in the same way and read, * a snb 
ewe,' 'a snb two," 1 etc., or merely ' a one,' l a two," 1 etc. 

SOLUTION. The same method of procedure is followed here as in 100, 1. 

(20 aa'x + ab'y = ac f ( (2) x a, Mult. Ax.) 

(li) aa'x + a 1 by = a f c ( (1) x ', Mult. Ax.) 

ab'y a' by = ac' a'c (Sub. Ax.) 

_ gc' a'c (Factoring left member, and Div 
~ ab'-a'b' Ax.) 



SIMULTANEOUS LITERAL EQUATIONS 



191 



Instead of substituting the value of y just found in one of the preceding 
equations to find the value of oj, it is just as easy to start again with the given 
equations and this time eliminate y. 

(li) ab'x + bb'y = b'c ( (1) x &', Mult. Ax.) 

(2i) a'bx + bb'y = be' ( (2) x 6, Mult. Ax.) 

(ab 1 - a'b}x --= b'c - be'. (Ax. ?) 

_ 6/c _ &c' 
'ab'-a'b 

Verification in (1) a . & ' c ~ &c ' + & . qc ' ~ q ' c = c, 

ab'-a'b ab'-a'b 

or, a&'c &c' + a&c' a'bc=ab'c a'bc. 

{mx + ny = p. faa? + &y = l. faic + &v = 2a&. 

9. J 10. J 11. 

c + c??/ = 1 . ( ox-\-ay = a~ o 

# + a?/ = 6. f ax by = 0. 

14. \ 



( mx nil = 0. 
12. \ 13. 



ax by = c. 



( ex dy = 1. 



15. ! 



16. 



a? a 



+l_a+l 
2/+1" 



17. 



a, 



y 



18. 



a; y 



SUGGESTION. Solve without clearing of fractions. 



19. 



22. 



c ^ see J lu, 25. 




L 1 ~~~ 




l# 2/ 




f- + - = c, 


a 1 T, 
- + - = 6, ( 




a? v 
20. \ 


21. 


:/-j /y 1 7^**?/ ^1 J 


m n 


5 






= a. 




T 11 

L jc y 


.a 2/ 




'x a a b 


flO 9 


' a? -(-_?/ a 




~i ~~~ ^) 




y a a-\-b 
23. 


U V 24. 


xy o c 


37 O 
- . /I* 5 
</ L/ 


2 3 


a; + c o + 6 


v a 3 -1- 6 3 




./ + & a + c 



192 LITERAL EQUATIONS 

120. Literal Equations containing more than Two Unknowns 

See 105, 106. 



1. y + a = 2x + 2z, 2. x -f- z y 6, 

[z + a = 3 x + 3 y. [y + z x = a. 

121. Problems involving the Solution of Literal Equations. 

1. A boy bought an equal number of apples, lemons, and 
oranges for c cents ; for the apples he gave I cents apiece, for the 
lemons m cents, and for the oranges n cents apiece. How many 
of each did he purchase ? 

SOLUTION. Let x = no. of each sort bought. 

Then Ix = no. of cents paid for apples, 
mx no. of cents paid for lemons, 
nx = no. of cents paid for oranges, 
and Ix + mx -f nx = c. 

(I + m -f n~)x = c. (Factoring.) 

x= - - -- (Ax.?) 
I + m+ n 

2. A person bought a certain number of lemons and twice as 
many oranges for a cents, the lemons costing I cents and the 
oranges m cents apiece. How many of each did he buy ? 

3. Divide b into two parts one of which is n times the other. 

4. Divide c cents between two boys so that one will have d 
cents more than the other. 

5. An estate of a dollars is divided between two heirs in the 
ratio of m to n. What is the share of each ? See 97, 16. 

6. Two men a miles apart travel towards each other, one m 
miles and the other n miles an hour. In how many hours will 
they meet ? If they travel in the same direction, the first behind 
the second, in how many hours will they be together? 

7. Divide the number d into three such parts that the second 
shall exceed the first by 6, and the third exceed the second by c. 



GENERALIZATION 1 93 

122. Generalization. The elementary algebra generalization of a 
problem is attained by replacing its known numbers by letters 
and then deriving its solution. Problems are studied using par- 
ticular numbers ; afterwards, on account of the frequency with 
which they occur, they are generalized. Then to work a problem 
one merely substitutes the particular numbers in formula found. 

123. Exercise in Generalization. 

1. Sum of two numbers is 256 and difference is 34. Find them. 

2. Generalize Ex. 1, replacing sum, 256, by s and 34 by d. 
SOLUTION'. Let x = first number and y = second number. 

Then, J* + y = s ; whence x = ^-^, y = 2-=^- ( 119, 1). 
[ x y = c I 2 2 

3. If the sum of two numbers is 375 and their difference is 33, 
find them by substituting in the formulas of Ex. 2. 

4. Generalize Ex. 1, 97, using a, b, c, d, e, for known numbers. 

SOLUTION. -+- + c = dx + e. Solving, x = ab ( e ~ c ) . 
a b a + b aba 

5. Solve Ex. 1, 97, by substituting in formula of Ex. 4. 

124. Indeterminate Equations. When there are more unknowns 
in a problem than given equations, the problem is indeterminate. 

125. Negative and Inconsistent Solutions. 

1. If from | of a certain number 1 is subtracted, the remainder 
equals f of the number. What is the number ? AUK. x = 8. 

The negative answer shows that there is some arithmetical 
inconsistency in the problem. Evidently if we take something 
from | of a positive number, we cannot have f of the number 
(which is greater than -f- of it) left. 

2. A father's age is 40 years, and his son's age is 13 ; in how 
many years will the father's age be 4 times that of the son ? 
Ana. 4 yr. 

This negative answer shows that the problem shoi Id have read 
"how many years ago" instead of "in how many years." 



194 SUPPLEMENTARY FACTORING 

* SUPPLEMENTARY FACTORING 

126. Cases in Factoring omitted in Chapter II. The more coin. 
plicated cases in factoring were not needed in. the preceding 
chapters, and were omitted from Chapter II. 

127. Factoring of Binomials Continued. See 51. 

I. Binomials that are the Difference of the Same Powers, or the 

Sum of the Same Odd Powers. 

By dividing, we learn that x 4 y 4 , x 5 y 5 , x e y e , etc., are all 
divisible by x y\ also, that or 5 + y 5 , x 7 + y 7 , etc., are divisible by 
x -f y. Thus, we get 

THEOREM IX. Tlie difference of the same integral powers of two 
quantities is divisible by the difference of the quantities. 

THEOREM X. The sum of the same odd powers of two quantities 
is always divisible by the sum of the quantities. 

1. Factor ce r -/; x 7 + y 7 ; 5 -326 5 ; x-z; ra 5 -f-243w 5 ; 
a 5 6 5 + 32c 5 ; a 7 -128; a 7 4-1. 

II. Binomials that are the Product of Two Trinomials. 

2. Factor a 4 + 4 b\ 

SOLUTION. (1. (a 4 + 4 aW + 4 6*) - 4 2 & 2 ( 116, 20). 

(2. (a 2 + 2 6 2 ) 2 - (2 a&) 2 . (Page 85, Ex. 30.) 

(3. [(a 2 + 2 5 2 ) + 2 ft][(a 2 + 2 & 2 )- 2 aft]. (Page 85, Ex. 30.) 

(4. la 2 + 2 ab + 2 & 2 ][a 2 _ 2 ab + 2 6 2 ]( 32, 13, 1). 

(5. Proof by actual multiplication. 

3. 4 +646 4 . 4. 4 a 4 4- 81 z 4 . 5. 64 ofy 4 -f- 81 z 4 , 

6. Make a rule for factoring binomials that are the product of 
two trinomials. 

7. 4m 4 + 625. 8. x 4 + 324,y 4 . 9. 4j> 4 +l. 



TRINOMIALS 

128. Factoring of Trinomials Continued. See 52. 
I. Trinomials that are the Product of Two Trinomials. 
Type form : mW paW + /? 2 6 4 . 



TRINOMIALS CONTINUED 195 



(1. (9 a 4 + 12 dW + 46*)- 16 a 2 & 2 . 

(Twice the product of the square roots of 9 a* and 4 6 4 is obtained and placed 
between these terms (see 116) ; then the last term is chosen so that 
the given trinomial is unchanged. Thus, 12 a 2 b 2 16 o 2 6 2 = 4 a 2 6 2 .) 

(2. (3 a 2 + 2 6 2 ) 2 - (4 a&) 2 . ( 52, II.) 

(3. [(3 a 2 + 2 & 2 ) -f 2 aft] [(3 a 2 + 2 6 2 ) - 2 a&]. (On factoring, 53, III.) 

(4. [3 a 2 + 2 ab + 2 6 2 ][3 a 2 - 2 a& + 2 6 2 ]. ( 32, 13, l.) 

(5. PROOF. By actual multiplication. 



2. 9 x 4 + 2 x 2 y 2 + y*> 3. z 

4. 25 a 4 - 36 aty 8 + 4 ?/ 4 . 5. 9 a 4 - 34 a 2 6 2 + 25 & 4 . 

6. Make a rule for factoring trinomials that are the product 
of two trinomials, giving steps. 

7. 9 a 4 + 26 cr& 2 + 25 b\ 8. 4a 4 -13a 2 
9. 16 x 4 - 17 x*f + 2/ 4 - 



SUGGESTION. This problem can be solved in two different ways : (1) by 
making the middle term of the trinomial square positive ; (2) by making it 
negative. The reason such a quantity can be solved in two ways can be 
illustrated in this way : 

60 = 2 x 2 x 3 x 5 = 4 x 15, or 6 x 10. 



10. 25a 4 -41zy + l62/ 4 . 11. 4a 4 -29a 2 a,' 2 -f 25 

12. 4a 4 -5a 2 l. 13. 25 a 4 + 71cr 



II. Trinomials, the Product of Binomials and Trinomials. 

14. 4 z 3 - 43 a- 21. 

SOLUTION. The trial method is used. 

2 x 2 + 7 x + 3 EXPLANATION. As in 52, III, the first coeffi- 

2 x 7 _ cients of the factors must give the first coefficient 

4 x 3 + 14 x 2 + 6 x of the given quantity, and the last coefficients of 

14 x 2 -- 49 x 21 the factors, the last coefficient of the given quan- 

4x 3 43 x 21 tity. The middle coefficient of the first factor 

must be so chosen that the x 2 term disappears in the product, and the middle 



196 SUPPLEMENTARY FACTORING 

term of the given trinomial is obtained. It may cause the student some 
trouble to find the desired coefficients ; but if he persists he will get a satis- 
factory combination, provided, of course, the given quantity can be factored. 



15. 9z + 5z + . 16. 

17. 8ar 5 -24z 2 + 25. 18. 21 x 3 + 26 or + 25. 

QUADRINOMIALS 

129. Factoring of Quadrinomials Continued. See 53. 

Quadrinomials the Product of a Binomial and Trinomial. 

m *~** ~ " ~ mxz 



Type forms : 

( ax 3 + 6jr 2 + ex + d. 

The first type is factored by putting two terras in one paren- 
thesis and the other two terms in another, and proceeding very 
much as in 53, II. The second type is factored by the trial 
method. 



(1. (4x 2 -25*/ 2 )-(4z2- 10?/2). (See 33.) 
(2. (4x 2 -25?/ 2 )-22(2z~5?/). (50.) 

(3. 2x-5?/|(4x 2 -25?/ 2 )-2 g (2x-6y) f c 50 29 ^) 

2x + 5y-2z 

(4. (2 x + 5 y - 2 2) (2 x - 5 y) = 4 cc 2 - 25 ?/ 2 - 4 xz + 10 ?/2. 

2 . m 2 -9rt 2 + 21ra?i + 63?? 2 . 3. 4 a 2 6 2 - 169 c 2 + 6 6d + 39 cd. 

4. a? 3 6 X 2 -j- 6 a? 1. SUGGESTION. Write as (x 3 -- 1) (6x 2 6 x). 

5. m 3 + 5m 2 + 5m + l. 6. 8 a 3 + 6 a 2 ?/ 2 -9 ay*-'27y & . 



7. 

SOLUTION. 2 x 2 + 3 x + 4 (See suggestions under 52, 73.) 

5x - 7 

- 14 x 2 21y-28 
10 r s + x 2 - x - 28 



POLYNOMIALS 197 

8. 4z 3 -21ar + 44z-30. 9. 6 or 3 -3 a 2 -33 a? -6. 

a. Quantities like the preceding and those at the end of 128 can often 
be factored by the method explained in 190, 70. 



POLYNOMIALS 

130. The Factoring of Polynomials of Five or More Terms. 

I. Powers of a Binomial. 
II. Square of a Polynomial. 

These quantities can be readily factored by the student when 
he becomes familiar with the forms these powers take. This 
subject of the powers of quantities will be studied in the chapter 
on Involution. 

III. Polynomials that are the Difference of Two Squares. 



(1. ( 2 + 2a&+ &2)_(4 C a_. i2cd+9eP). ( 52, II, 33.) 

(2. (a + &) 2 - (2 c - 3 dy 2 . 

(3. [(+ &) + (2c-3d)][(a + 6)--(2c- 

(4. [a + 6 + 2 c - 3 d] [a + 6 -- 2 c + 3 d]. 

(5. Proof by actual multiplication. 

2. aj 2 --2 xy + y 2 - - m 2 + 2 mn--n 2 . 

3. x~-2 

4. 4 m 2 - - 12 mn + 9 n 2 -x? - kpq - - 4 q 2 . 

5. (a + b + c) 2 rf 2 . SUGGESTION. Factor as J. 2 B 2 . 
6. 



7. (2 a& + 2 erf) 2 - (a 2 + 6 2 - c 2 - d 2 ) 2 . 
SUGGESTION. After factoring, each factor can be factored. 

8. 4a; 



198 SUPPLEMENTARY FACTORING 

IV. Polynomials Separable into Two Trinomials or a Trinomial 

and Binomial. 

9. 6x 2 llxy + 3y 2 xz + 5yz-- 2 z 2 . 
SOLUTION. We will use the trial method. 

3x y 2z 
2x By + g 



6 x 2 - 2 xy 4 xz 
- 9xy 

+ 3xz 



6 x 2 11 xy - xz + 3 y 2 + 5 ye 2 z 2 

EXPLANATION. The coefficients of x in the factors are chosen to give 6 ; 
the coefficients of y to give 3 ; the coefficients of z to give 2. If the factors 
selected do not give the intermediate terms of the given quantity, another 
combination must be chosen. 



10. 2 

11. 6a 

12. a* 

13. 2a 

14. 2a 

15. x 2 

SOLUTION. (1. (x 2 2 xy + ?/ 2 ) + (5 x 5 /) 
(2. (x-y) 2 + 5(05-0). 
(3. (a-lOLXs-lO+6]. 

16. # 2 + xy - - 12 y 2 + 2 o;z + 8 ?/z. 

17. 3ic 2 + 2a;?/--4/ 2 9a; 2 



131. Miscellaneous Exercise in Fractions involving the use of 
the preceding cases in factoring and 60. Simplify the quanti- 
ties, and check, if testing is desired, with figures. 



m 4 



- 

' 



6a; 9 oj 8 oj 6 

4. ? 



MISCELLANEOUS EXERCISE IN FRACTIONS 



199 



7. 



10. 



12 









4 x 4 - -4 a- 2 // 2 + 9?/ 4 



11. 



13 



6-fc-a ' 






14. 



15. 



16. 



17. 



18. 



O O f~\ 

x 2 -- xy xz 



a? 



1- 



x 



(a? - ? /) 2 - (5 - a) 2 . 



(a. + a)* -(y + b) 2 (x - a) 2 - (6 - y)* 



_(a - b) 



c 



x 



c 



frfZ I ,,,2 ^,2\2 

I *C' ' ~ */ ~ ^ ) 

4 ?/ 2 



19. (12 OJ 2 + 5a?/ - 2 



2 xz 






2 a; 4 5 2 



132. Historical Notes. Algebra got its name in Arabia. The 
full title of the first Arabic work on algebra was " Al-gebr we'l 
mukabala." It means that the same quantity may be added to or 
subtracted from both sides of an equation. (AL means the, as in 
other Arabic words, such as almanac, alchemy.) Thus, the 
thought in the minds of the first algebraists was that algebra is 
the axiomatic science. 



200 BIOGRAPHICAL NOTES 

The equations of 124 are often called Diophantine equations, 
after Diophantus, the early writer on algebra ( 43). It is a 
curious fact that Diophantus spent more time in solving these 
indeterminate equations than in solving the determinate equa- 
tions in which other writers have been most interested. 

When Descartes invented the method of constructing graphs 
from equations (see 114), he is said to have applied algebra 
to geometry. Thereafter these two subjects, which had before 
been separated, became, so to speak, married. The analytical 
geometry of Descartes, an extension of the study of graphs, is a 
powerful instrument for the study of geometry, just as algebra 
enables us to solve more difficult problems than those of 
arithmetic. 

133. Biographical Notes. It may be worth while for the 
student to interest himself a little in the mathematical studies 
of some of the world's eminent men. Short biographies usually 
pass over with a few words the education of the person described, 
but the larger ones often contain interesting details. 

Washington had no training in languages, but considerable in 
mathematics. In his earlier years he had a little schooling in 
reading, writing, and arithmetic. When fourteen years of age 
he wanted to go to sea, but his mother at the last moment 
failed to give her consent. He then attended a school kept by 
Mr. Williams at Bridges Creek, Virginia, where he continued for 
two years, paying special attention to mathematics, as he intended 
to fit himself for a surveyor. There are still extant some frag- 
ments of his school exercises which show that his work was neat 
and accurate, and that he tried to master his subjects. After he 
left school, and when at his brother Lawrence's and at Lord 
Fairfax's, he kept on with his mathematical studies. The alge- 
bras in current use in England at the time, one or more of which 
he probably studied, were Harriot's, Oughtred's, Wallis's, and Sir 
Isaac Newton's (Universal Arithmetic). 

Jefferson also was strong in mathematics. His biographer says : 
" Jefferson was at his best in mathematics, and he could read off 



BIOGRAPHICAL NOTES 201 

the most abstruse processes with the facility of common dis- 
course. This study he kept up as long as he lived ; and he 
delighted in applying its principles to anything and everything." 

How Lincoln came to study mathematics beyond the arithme- 
tic he learned in early life is an interesting story. In Nicolay 
and Hay's life of Lincoln we find this passage: "It was at this 
time that he gave a notable proof of his unusual powers of men- 
tal discipline. His wider knowledge of men and things, acquired 
by contact with the great world, had shown him a certain lack in 
himself of the power of close and sustained reasoning. To 
remedy this defect, he applied himself, after his return from 
Congress, to such works upon logic and mathematics as he fan- 
cied would be serviceable. Devoting himself with dogged energy 
to the task in hand, he soon learned by heart six books of the 
propositions of Euclid, and he retained through life an intimate 
knowledge of the principles they contain." 

General Grant was good in mathematics and at one time hoped 
to get a detail as assistant in mathematics at West Point. 
" Grant," says a schoolmate of his, " was strong in mathematics. 
I studied algebra with him, and I remember that he would never 
let Carr White or me show him the way to do problems, but 
always wanted to work them out himself. He had a way of 
solving problems out of rule by the application of good hard 
sense." In one of his letters home from West Point he speaks 
of having long and hard algebra lessons to get. 

Napoleon Bonaparte was distinguished for his application to 
the study of mathematics and took prizes in it. 

Peel and Gladstone, two of England's greatest prime ministers 
in the nineteenth century, took double firsts in languages and 
mathematics when they finished at Oxford University. 

The seven characters just named stand among the greatest of 
all the men of modern times, and we have seen each of them 
interested in mathematics and some of them masters of the sub- 
ject. If the lives of lesser yet successful men are read, some- 
thing like the same facts will be found true of many of them. 



CHAPTER VIII 

INVOLUTION AND EVOLUTION 

I. INVOLUTION 

134. Involution, as a term in algebra, refers to raising quantities 
to powers. An integral power of a quantity may be denned as a 
continued product arising from multiplying unity by the given 
quantity as many times as the exponent of the power indicates. 

Thus, a 3 = lxaxaxa. 

135. Raising Monomials to Powers. 

Type form : (xy b ) n = jf a y bn . 

EXAMPLES. (3 m*n 8 ) 2 = 1x3 m 2 w 8 x 3 m 2 n 3 = 9 m*n & . 

(-2a 3 ) 3 = lx -2a 3 x -2a 3 x -2a 3 =-8a 9 . 

1. Square 4 a 2 , 3 a 3 , - 2 m\ 5 x 2 y\ - 11 b s c 2 , - 4 xy\ -Sy 5 . 

Check results in the preceding and in the following problems 
whenever there is any doubt of their correctness. Thus, to test 
(4 a 2 ) 2 = 16 a 4 , put a = 2. Then, 4 a 2 = 16, and 16 2 = 256 ; also the 
answer, 16 a 4 = 16 x 16 = 256. 

2. Cube2aa;; 3 a 2 ; -6 3 d; c 4 ; 5cZ 5 ; -36 2 c 3 ; -7 xy z ; -4 arty 6 . 

3. Raise to 4th power a 2 ; 3m 2 ?i; 2a 4 6 3 ; -c 5 d 7 ; -2 a 12 . 

4. Raise to 5th power - 2 a; - 3 6 3 ; - 2 c 2 d 4 ; - c 7 c? n ; - m 6 . 

5. O 3 ) 5 ; 0V) 6 ; (-3 a;?/ 2 ) 4 ; (-2mV) 3 ; (-6 2 2/V) 2 ; (-abc) 

5 o\ 3 / 2\ 5 



202 



5 



RAISING MONOMIALS TO POWERS 203 

7. Make a rule for raising monomials to powers. 

(1) What sign does the power always have if the given quan- 

tity is positive ? \Vhat sign has the power if a negative 
quantity is raised to an even power ? To an odd power ? 

(2) How is the coefficient of the power found ? 

(3) How is the exponent of a quantity in the power found from 

the exponents it has in the given problem ? 
Ans. Multiply the exponent of the quantity by the index of the 
power for the exponent of the quantity in the answer. 

(4) State how the result can be checked for accuracy. 

8(9, r/ 2 r a ?/ 4 V ( 2 T- 2 */ 3 ?^ 3 ( X nh 2 r 3 \ s ( - 10 o^ 3 > ( 1 "\99 
. I *J (Aj JU U J . I ~J JU fj 6 I j I tJttC/U) j I -LV*t/) . I J_ ) . 

9\ 9 f rt \ 1 /r\ 4\ Q / Q <! \ R 

o*^\ / /, n\ / M o)? )> \ c ' / n"vn \ 

JU \ I fj \Ai \ I U III U \ I Ui I Ik \ 



o / ' I ^2,/'\ o s J'lova J' 

o y v or?/y \ 



9. ( a# m ) n = ? Show that the solution of this problem gives 
a general formula or rule for solving any problem. Regard a as 
the coefficient of x m . If n is odd and the given sign is , what 
sign has the power ? Otherwise, what sign has it ? 

10. Cube 3 a 4 ; - - b 4 y 2 ; a 13 x a 10 ; 2 a n ; a y ; raise Wy to the fourth 
power ; c 3 x to the fifth power ; 2 m s n 2 to the sixth power. 



11. 



. /2a&cV. 



12. The student should now turn back to 27 and study that 
article anew. In 2 a 2 + 5 (3 a 2 6 2 ) 3 -*- 4 a 4 6 6 , which operation must be 
performed first ? Which comes next ? Which next ? Which last ? 

13. Simplify 3 <26 2 m 3 ) 3 ; 2a(4a 2 ) 2 ; 4(-4opV') 2 ; 
(_3n) 4 x4(-2cm 4 ) 3 ; a 



14. Simplify 5(3 a) 4 (2 6) 2 ; 10 (f a) 3 (f) 2 ; 9(3 a 2 ) 4 ( - 2 b 2 ) 3 ; 
12(2 a:) 3 (- y) 3 ; T ?,(4 m 3 ^.) 4 (3 n) 3 ; 10(2 z) 4 (3 y) ; Kl 3 



204 INVOLUTION AND EVOLUTION 

136. Raising Binomials to Powers. Newton's Theorem. 

1. By actual multiplication show that 
(A - ) 2 = A 2 - 2AB + B 2 . 
(A - BY = A* - 3 A 2 B + 3 AB 2 - B 3 . 
(A - B)* = A* - 4 A*B + 6 A*B* - 4 AB* + B*. 
(A - ) 5 = A* - 5 A*B + 10 A*B 2 - 10 A*B* + 5 AB* - B 5 . 

- 20 A 3 B* + 1 5 A*B* - 6 .4 5 + B*. 



Examining eacft of these products we find that 

(1) If A B is raised to any power, the signs of the terms are 

alternately positive and negative. Evidently if A -f- B is 
raised to any power, all the signs are positive. 

(2) The exponent of A in the first term of the right side is always 

the same as the exponent of the binomial on the left side. 
How are the exponents of A in the remaining terms always 
found ? Which is the first term containing B on the right side 
in each case ? How are the remaining exponents of B found ? 

(3) What is the coefficient of the second term on the right side each 

time as compared with the exponent of the binomial on the 
left side ? 

(4) The student will not readily see how the remaining co- 

efficients may be found. They can always be obtained in 
this way : Each coefficient can be found from the preceding 
term by multiplying its coefficient by the exponent of its leading 
letter and dividing the product by one more than the exponent 
of its other letter. 
Thus, 10 in 10 A S B 2 in the value of (A - ) 5 comes from 

tMis : : - = 10, where 2 is one more than the exponent of B understood. 

12 

2. Expand by Newton's theorem (a b) 4 . 



SOLUTION, (a - 6)* = a* -- 4 a 3 & + a% 2 - - a& 3 + 



= a 4 - 4 a 3 6 + 6 a-b 2 - 4 ab* + & 4 . 

CHECK. Let a - 3, 6 == 1 . Then, (3 -- 1) 4 = 2 4 == 1C ; also 

- 4 aW + 6 4 = 81 - 108 + 54 - 12 + 1 = 16. 



RAISING BINOMIALS TO POWERS 205 

3. Expand (a b) 5 by Newton's binomial theorem and compare 
the result with that given on p. 204 in Ex. 1. Similarly expand 
(x y) 6 and compare the coefficients and exponents with those 
given on p. 204. 

Expand in the following problems and check as in Ex. 2 until 
';he process of checking is well understood. 

4. (a + 6) 4 ; (a-6) 7 ; (x + yf; (w+w) 9 ; (p-q) w . 

5. Develop (3 m 2 2 n) 4 by Xewton's binomial theorem. 

SOLUTION. Notice that 3 m 2 takes the place of A and 2 n that of B. To 
preserve the exponents of A and B for use in getting coefficients it is neces- 
sary to write 3 m 2 and 2 n in parentheses. 

(3 w 2 - 2 7i)4 = (3 ?n 2 ) 4 - 4 (3 m 2 ) 3 (2 n) + ^-^ (3 m 2 ) 2 (2 w) 2 ^^ (3 m) (2 ) + (2 n)* 

^ o 



= 81 w8 216 m*n + 216 m*n* 96 m%8+16 n 4 . ( 27) 

To check the result let m = 2,n = l. Then, (3 m 2 - 2 )* = 10,000; also 
81 m 8 - 216 w% + 216 w 4 2 - 96ra 2 n 3 + 16 w* 

= 20736 -- 13824 +3456 - 384 + 16 = 10000. 

a. In the preceding solution we think of (2 m 2 3 n) as written 
[2m 2 (+ 3n)]*. If we wrote it [2w 2 + ( 3 )] all the signs between 
terms would be +, and 3n instead of + 3 n would appear in parenthesis 
each time. The results would be the same. 



6. (2a + 56) 4 . 7. (2.T + 3?/) 6 . 8. 

9. (4a 2 c 2 +5c 3 ) 3 . 10. (3a 2 -56 2 ) 4 . 11. (a 2 - 

12. (2 a 2 b - 7 c 3 ) 4 . 13. (9 a 7 - 4 /) 3 . 14. 



b. Let the student write clown the values of (a + &) 2 , (a + ft) 3 , (a -f 6) 4 , 
(a + &) 5 , (a + &) 6 , and memorize the binomial coefficients, so that he will 
not have to calculate them each time. Thus, the binomial coefficients for the 
fourth power are 1, 4, 6, 4, 1. 

15. (5 -4 a?) 4 . 16. (Go 3 -/) 4 . 17. (2 a 2 -I) 8 . 

NOTE. When either term is 1, as in Ex. 17, retain the 1 in parentheses 
with its different exponents, using these to get coefficients by Ex. 1, (4), p. 204 



206 INVOLUTION AND EVOLUTION 

3 2x\ 6 



18. 



9 > 



SUGGESTION. - - -6- - + 16 - -etc. 



3\4/O r \2 

; -) (^ ) - 

2a; / V 3 / 



- etc. 



729_243 + 135_ etc< 



64 x 6 8 x 4 4 x 2 



19. x-. 20. z-S. 21. * *- 

ar 2/ 



22. (Qp-tf 1 )*. 23. (l + a n ) 5 . 24. 

25. Find first four terms of (a + b) 15 ; of (x y) 20 . 

26. Calculate 99 3 by writing it (100 I) 3 . Check by actual 
multiplication. 

SOLUTION. (100 - I) 3 = 100 3 - 3 (100) 2 + 3 (100) - 1 

= 1000000 - 30000 + 300 - 1 = 970299. 

27. Calculate in the same way 998 3 =(1000-2) 3 ; 197 4 ; 
(12i) 4 = (12 + i) 4 ; (29i) 2 ; 389 4 ; 205 4 ; 1010 3 . 



137. Raising Polynomials to Powers. This is done by changing 
the polynomials to binomials by inclosing terms in parentheses. 

1. (x 2 -x+2) 3 . 

SOLUTION. Change given trinomial to binomial. 
[(a;2 _ X ) + 2] = (x 2 - x) 3 + 3 (x 2 - x) 2 (2) + 3 (x 2 - x)(2) 2 + & 
= ( X 2)3 _ 3 ( X 2)2(a;) + 3 (x 2 ) (x) 2 - x 3 + 3[x 4 - 2 x 3 +z' 2 ] x 2+3[4 x 2 -4 x] +8 
= x 6 - 3 x 5 + 3 x 4 - x 3 + 6 x 4 -- 12 x 3 + 6 x 2 + 12 x 2 - 12 x + 8 
= x 6 - 3x 6 4- 9 x 4 - 13 x 3 + 18 x 2 - 12 x + 8. Check with x = 1. 



EXTRACTING ROOTS OF MONOMIALS 207 



2 . (l-a + or 9 ) 3 . 3. (m 4 -n 2 -3) 3 . 

4. (a + 6 + c) 4 . 5. (2a 2 -3a-4) 3 . 

6. (a -1- b + c + d)\ SUGGESTION. Write thus [(a -f 6) + (c + d)] 4 . 

7. (a-6 2 + 2c-3cf) 3 . 8. (i-f 2n- 
9. (1 + a + ary. 10. (2 cT - b n 

11. 1799 3 = (1000 + 800 I) 3 . Verify answer by multiplication. 

12. (a + 5 + c) 2 . 13. (a + & + c + d) 2 . 

14. By generalizing from the results of Ex. 12 and 13 we get 
this theorem: The square of the sum of any number of quantities 
equals the sum of their squares plus the sum of twice the product 
of each quantity by each quantity that follows it. 

Write directly the answer in the following by using the theo- 
rem just enunciated. Check Ex. 15, 16. 

15. (x 2 -3a + 2) 2 . 16. (xy + yz+zx) 2 . 
17. (a 3 -- a 2 -- a -- 3) 2 . 18. (2 a 8 - -4 x 2 - -3 
19. (15 a - 9 y - 12 z) a . 20. (3m + n 

II. EVOLUTION 

138. Evolution in Algebra means the extraction of roots of 
quantities. A root of a quantity is one of its equal factors ; or 
it is a divisor which being divided into the given quantity and then 
into the quotient, and so on, gives the quotient unity. 

139. Extracting Roots of Monomials. 



Type form : ^/a n b n = ab. 

1. What sign should be given to the square root of a number, 
as 4 or 9 ? Prove 2 2 = 4 ; also that (-- 2) 2 = 4. What sign, then, 
should be placed in front of the root? (See 36, a.) What 
sign should the cube root of 8 or 27 have ? What sign should the 
cube root of 8 or 27 have ? What sign should the fifth root 



208 INVOLUTION AND EVOLUTION 

of 32 have ? The fourth root of 16 ? Ans. to last question . 
Prove the answers correct in each case. 

What sign should be given to an odd root of a positive quan- 
tity ? To an odd root of a negative quantity ? To an even root 
of a positive quantity ? 



2. Show that V- -4 is neither + 2 nor 2. The square root 
of a minus quantity is commonly called imaginary. 

3. Write the square root of 36a 2 6 2 ; of 9aW; of 



16 

4. -v/ 27aV. 5. A/ -8 x 6 u 12 . 6. -J/16 a 4 6 8 c 



7. A/ -64 a 12 . 8. -x/lGa 5 . 9. ^/1728 a 12 ?/ 3 - 

10. V-16a 4 . 11. V2W. 12. A/-243a u V 5 . 



13. 125a 21 . 14. tf. 15. - 



/ 27 a^v'" 
' \- -64^' 



25 c 4 V'a (i! 6 14 \ 64 

19. How is the coefficient of the root found from the given 
coefficient ? How is the exponent of a quantity in the root found 
from the given exponent of the quantity and the index of the 
root? 



20. J/^ri*. 21. VlSV 1 . 22. -v / -2 10 a 5 6 10 c 20 . 



23. V81x 64x121. 24. V(64a 8 ) 2 . 25. 



481 a? 128 4181 12 " 1 

26 ' 



Simplify in following problems, taking only + value of even 
roots : 

29. V25 a 2 6 4 c 2 + A/' - 8 aW - -\/81 a W - >/ - 32 a 



30. V - 27 ar 3 / x V 243 ?/V X Vl6 



SQUARE ROOT OF POLYNOMIALS 



209 



140. Square Root of Polynomials. 

1. The method of extracting such roots is obtained from a 
study of the familiar formula/ 



It is required to find a process to obtain A + JB from A 2 + 2 AB -f B' 2 . 

A + B 



A 2 + 2 AB -f B 2 
A 2 



2A+ B 



2 AB + B 2 

2 ^45 -} B 2 



derneath the line obtained ? 



SOLUTION. How is A in the root found 
from A 2 + 2 AB + B 2 ? From which term 
is it obtained and how? ( 116.) How is 
the A 2 under A 2 obtained ? Ans. By squar- 
ing A in the root. How is 2 AB -f B 2 un- 
How is 2 A (called the trial divisor) , in the 



divisor at the left, obtained from A in the root? Ans. By doubling A 
( 116). With the divisor 2^1 and the dividend 2 AB + B 2 , how is Bin 
the quotient found ? Where else is the quotient of 2 AB -^-2A written be- 
sides in the root ? Ans. At the right of the divisor 2 A . 2 A + .B is called 
the complete divisor. How is the operation completed ? 

2. Extract the square root of 



SOLUTION. Arranging this ( 37) with reference to the powers of x (and 
y and 2), we have 



- 4 



4- 12 xz* 



- 6 



~ ay+ 3z* 



4 x ay 

4r+ fc 
v * 


4 /7^"?/ 1 1 9 O" ?*2 1 /y2?/2 
^1 LtiX^y ^^ A *il/v \ \Jv ti 

4 r?/ + -y 2 


?-4 
-' ^ 


5 , + 3, 2 


12 xz 2 - 6 a^ 2 4- 
12^ 2 -6^ 2 + 



How is 2 # in the root found ? How is 4 x in the first divisor found ? How 
is ay in the root found ? In what two places is ay written when found ? 
What are the next operations ? How is 4 x 2 ay found ? By dividing what 
by what is 3 z* found ? Where is. 3 z' 2 written besides in the root ? What is 
last operation ? How can result be proved correct ? Ans. By squaring the 
result. (See 137, 14. ) The answer to this problem is (2 x - ay + 3 z' 2 ) 
since squaring either 2 x ay + 3 2 2 or ay 2 x -- 3 2 2 produces given poly- 
nomial. Is it important to arrange the terms in a given problem and also 
the remainders as they are obtained? ( 37, 38.) 



210 INVOLUTION AND EVOLUTION 

3. a 4 + 6or ) -h9 2 . 4. 49 a 4 42 a?b + 9 fc 2 . 

5. a! 4 - 

6. 4ce 8 



7. 25 a 4 - 30 oaj 8 + 49 ah? - 24 a B x + 16 a 4 . 



8. l_ 



9. a;- 
10. 

a 4 a 2 6 3 6 6 a 2 c 4 c 8 



12. 

13. !- 

14. a? 6 + 4 ax 5 10 a V + 4 a 5 a? + a 6 . 

In the two following problems extract the 4th root by extract 
ing square root twice : 

15. ^(16 x 4 - 96 tfy + 216 x 2 y 2 - 216 xy 5 + 81 ?/ 4 ). 

16. ^/(a 4 + 8 o 8 6 + 24 a 2 6 2 + 32 a6 8 + 16 5 4 ). 

17. 6 ab 2 c - 4 a 2 bc + a 2 & 2 + 4 a 2 c 2 + 9 6 2 c 2 - 12 

18. 5 

19. Write a rule for extracting the square root of polynomials. 

20. l- 



JL.L + - ^.i^L 8 . i^- + 
' 25 25 36 25 5 49 35 15 

4 g 4 ?i 4 2 m 3 ? 
~35~ ~2T 



SQUARE ROOT OF ARITHMETICAL NUMBERS 



211 



22. Obtain three terms of the square root of 

/y.2 /v4 



SOLUTION. 



a* + x 2 



2a + - 
2a 



2a + ^- 



r 2 _L 



a 8 a* 



X 4 



X 4 



_ _^_ +_ 
n 



a 2 8 a 4 64 a* 



PROOF. 



8 a* 64 s 

x3. (137,14.) 



2a 8aV 8 

23. Obtain three terms of the square root of a?b 2 c 2 . Prove. 

24. Obtain three terms of the square root of 9 a 4 4 & 4 . Prove. 

In the following get the square root three times in succession : 

25. -fy(x 8 + 8x 7 + 2Sx & + 56x : + 70x* + 56x 3 + 2Stf + 8x + l). 

26. ^(266 a 8 + 3072 aj r + 16128 a 6 + 48384 x 5 + 90720 a 4 

+ 108864 a 3 -f 81648 aj 2 +34992 a;+6561). 

141. Extraction of the Square Root of Arithmetical Numbers. 

1. Extract the square root of 625. 

Since 625 lies between 400 and 900, its square root lies between 20 and 30, 
that is, between 2 tens and 3 tens. For guidance in the solution we use the 
formula ^ + u y = t z + 2tu+ U 2 = p + (2 t + ?Q u 

in which t stands for number in tens' order and u for number in units' order. 
The process at the right below is just like that at the left, which was 
explained in the last article. t + u 

t 2 + 2tu + u 2 \t + u 6'25 120 

t 2 P = 4 00 



u 



Trial divisor = 2 t = 40 



u = 5 



2 tu + w 2 



Complete divisor = 2 t + u = 45 



225 



225 



212 



INVOLUTION AND EVOLUTION 



If a number has an even number of integral orders, or none, each pair 
of its figures gives one figure in the square root : if it has an odd number of 
such orders, its first figure gives the first figure in the root; and after that 
each pair of figures in the number gives one figure in the root. For this 
reason, in extracting the square root, numbers are separated by marks into 
" periods " of two figures each commencing at the decimal point. 

In the following solutions ciphers not actually needed are not written. 
Instead of writing 40 + 5, as on p. 211, it is enough to write 4 only and then 
annex 5 to it when 5 is found. 



2. 3445.69. 
SOLUTION. 



3. .0003627. 



34'45'.69[58.7 
25 



SOLUTION. 



'.OQ'03'62'70[. 01904+ 
.0001 




1167 



29 

38( 


262 
261 


)4 


1 70'00 
15216 



1784 



Ex. 2. Into periods of how many figures is the number pointed off ? 
At what point does one start in pointing off ? How is 5, the first figure of 
the root, found ? Why is it not 6 or 4 ? How is 25 found ? 10 in the divisor 
at the left ? 8 in the root ? Where is 8 placed besides in the root ? Ans. 
At the right of 10. How is 864 found ? 116 ? 7? 8169 ? 

Ex. 3. What is the first thing done? Where does one start to point off'? 
Why is the cipher annexed to 7 ? What does the first period of two ciphers 
give in the root ? When the cipher after 9 is placed in root, where else is it 
placed at the same time ? Remember that the trial divisor in square root 
always has understood. 

PROOF. By actual multiplication (58. 7) 2 = 3445.69 ; 
also (.01904) 2 + .0000001784 = .00036270. 



4. 484. 

7. 4624. 

10. 494209. 

13. 74.1321. 



5. 1024. 

8. 18769. 

11. 328329. 

14. 60.3729. 



6. 7396. 

9. 106929. 

12. .165649. 

15. 4076.8225. 



16. .004761, 



17. 150.0625. 



18. .00459684. 



R 



S' 



CUBE ROOT OF ALGEBRAICAL QUANTITIES 213 

Find answer in Ex. 19-26 to three decimal places, Ex. 27-31 
to four, and Ex. 32, 33 to five places. If the fourth figure in Ex. 
19-26 is one of the figures 5-9, then in giving the answer the 
third figure should be made one greater. In proving the problem 
keep last figure unchanged, square, and add remainder. 

19. 3. 20. 5. 21. 7 T 4 T . 22. 2. 23. 10. 

24. 8J. 25. 0.3. 26. 0.5. 27. .0021. 28. .723. 

29. .062. 30. .000067. 31. f. 32. f. 33. -fa. 

34. To explain square root by means of a geometrical figure, 

let us take 729 and extract its square root. 

_ ? 

Suppose there are 729 little square units in a 
given square. Find the number of linear units on 
one side of the square. 

How is 20, one side of, the square S in the dia- 
gram, found ? Why is one side of it not 30 ? When 
S is found, what is the length of rectangle R and 
of R ? How many square units are left out of the 
729 to be put around 8 into R, R', and S'? Ans. 
329. Neglecting S' as small compared with R + 

R' + S', how can the width of R + R> be found from their total length, 2 x 20, 
and their area 329- ? A ns. By dividing 329- by 40. (329- means some- 
thing less than 329.) What is 2 x 20 called in the process of extracting the 
square root? Ans. Trial divisor. If the length of #' (found by dividing 
329- by 40) is added to R + R' , what is the total length of R + R + S'? 
What is this 47 called ? Ans. Complete divisor. 

35. Explain geometrically the extraction of the square root of 
576; of 3364; of 6068.41. 

*142. Extraction of the Cube Root of Algebraical Quantities. 
1. By Newton's theorem (see also 45, V), 



20 



R' 



20 



* See statement about starred articles on p. 6. 



214 



INVOLUTION AND EVOLUTION 



It is required to frame a process to obtain A + B from A* + 3 A 2 B + 3 AB* 



A* 



3 A 2 B + 3 AB 2 + 



How is A the first term of the root found ? How is 3 A 2 B + 3 AB 2 + B" 
on the third line found? How is 3 A 2 (called trial divisor) found from 
the first term of the root ? Ans. By squaring the first term and multiply- 
ing the result by 3. How is B, the second term of the root, obtained from 
3 A 2 and the first term of the remainder, 3A 2 B? Tell how 3AB + B 2 
(what is added to 3 A 2 for the complete divisor) is constructed from A and 
B, the first and second terms of the root. Ans. To the trial divisor .are 
added three times the product of the first term of the root by the secondhand 
the square of the second. What operation is performed last ? 

2. 



-3Qx 5 + 54**- 27 



+25 



60z 4 -180z 3 +285;c 2 -225+125 



60 a 4 - 180 x 3 + 285 y? - 225 x + 125 



EXPLANATION. The first 12 se* = 3(2 z 2 ) 2 ; the - 18 x 3 = 3 x 2 yt x - 3 a; ; 
x 2 = (-3^) 2 ; underneath, the 12 x* 3Gx 3 + 27ic 2 = 3(2x 2 -3x) 2 j etc. 
Apply the questions under 1 above, to this problem. 

3. 8w 3 -60m 2 >i-r-150m^ 2 -125w 3 . 

4. 27 a 8 ^ - 108 a 2 6 2 c + 144 abc* - 64 c 3 . 

5. a 6 - 

6. a 

7. 6 + 12 a^ + 63 a 4 + 184 a^ + 315 a 2 + 300 a; + 125. 

8. 125 x 9 - 300 x? + 465 a; 7 - 424 a 6 + 279 x 5 - 108 x* + 27 a 3 . 

9. a^ + 6aS* + 15a? + 20 + i5 + ^+i. 

or or ic 6 

1 0. 60 cV + 48 cz 5 - 27 b 6 + 108 <?x - 90 c% 2 + 8 x 6 - 



CUBE ROOT OF ARITHMETICAL NUMBERS 



215 



11. Make a rule for extracting the cube root of algebraical 
quantities. See questions under 1, p. 214. 

12. 28 a?W - 3 a~b 4 - 54 a 5 b + 27 a 6 + 9 a*b 2 - W - 6 ab 5 . 

Find the sixth root of the following two exercises by extracting 
the cube root first and then the square root : 

13. a 12 - 6 a w b + 15 a s b 2 - 20 a fi b 3 + 15 a*b* - 6 a 2 b 5 + W. 

14. x l2 -12 x ll y 2 + 60 x l Y-l6Q x 9 f +240 ^-192 a?V+64 a 6 ?/ 12 . 

15. Extract cube root of 8 a 3 W to three terms (see 140, 22). 

16. Extract cube root of 1 a? to three terms. 

17. a s^ 



*143. Extraction of the Cube Root of Arithmetical Numbers. 
1. Extract the cube root of 19683. 

For guidance in the solution we use the formula 

(t + w) 8 = t s + 3 Pu + 3 tip + w 3 



in which t stands for tens' number and u for units' number. 

In square root we saw numbers divided otf into periods of two figures 
each. In the process of extracting cube root the given number is divided off 
into periods of three figures each, commencing at the decimal point, be- 
cause each period of the number gives one figure of the root. The last 
decimal period should be filled out, if not full, with ciphers. 



SOLUTION TO EXPLAIN PROCESS 

t +U 

19'683 20 + 7 
*3 = 8 000 



MODEL SOLUTION 
SAME ABRIDGED 
19'683 |_27 
8 



Trial divisor, 3 P = 1200 
+ 3w= 420 
+ 11*= 49 



Complete divisor = 1669 



11683 



11 683 



1200 

420 

49 



1660 



11683 



11683 



Should a root contain three figures, as 275, after the first two figures are 
found, a cipher must be annexed to 27 in getting the trial divisor and the 
3tu term, since 270 is now regarded as t in the formula, and 5 as u. 



216 INVOLUTION AND EVOLUTION 

2. Extract the cube root of 158.252632929. 

158'.252'632'929 | 5.409 EXPLANATION. Into periods of how 
125 many figures each is the number pointed 



7500 

600 

16 



8110 



33 252 ^ ky marks ? From what point does this 

marking start ? How is 5 in the root 
found ? Why is it not 6 ? Why not 4 ? 

How is 7500 found ? 4 ? 600 ? 16 ? 8116 ? 
32 4^4 



What two operations are performed next 
874800 00 788 632 929 one after the other ? How many figures 

1458 00 are brought down at one time after the 

81 remainder is found ? 

876258 81 788 632 929 How is 874800 found ? Ans. By squar- 

ing 54 regarded as tens ; that is, 540, 

and multiplying the result by 3 as called for in the formula term 3 f 2 . 
When 874800 is not contained in 788632, what is done ? Ans. One cipher 
is written in the root, and two at the right of the trial divisor and a new 
period of three figures is brought down. How is 145800 found ? How is 
81 found ? 

When is the decimal point inserted in the root ? Is it necessary to retain 
the decimal point in the divisors and remainders ? How can the answer be 
checked? 

As a preparation for working in cube root the student should memorize 
I 3 = 1, & = 8, 33 = 27, 4 3 = 64, 5 3 = 125, 6 3 = 216, 7 3 = 343, 8 3 = 512, 9 3 = 729. 

3. 12167. 4. 74088. 5. 300763. 

6. 941192. 7. 681472. 8. 8741.816. 

9. .059319. 10. 175.616. 11. 410.172407. 

Find the following to three decimals and in last two instances 
to four decimals. Prove by actual multiplication, and adding 
remainder : 

12. 2. 13. 3f. 14. 100. 15. 500. 16. .6. 

17. .35. 18. f. 19. f. 20. f. 21. f\. 

22. Explain cube root by means of the " cube root blocks," 
drawings of which are given, p. 217. The blocks themselves are 
inexpensive and should be used in the explanation. 



CUBE ROOT OF ARITHMETICAL NUMBERS 



217 



Let it be required to extract the cube root of 13824 by following a geomet- 
rical explanation. To do this the 13824 is regarded as 13824 little cubes in 
one large cube (lowest right hand figure; and it is required to find the 
number of little cubes along one edge of the large cube. 





F 



E 






3 x 20 2 = 1200 

S x 4 x 20 = 240 

4 2 = 16 



1456 



13824 1 20 
8000 



5824 



5824 



, EXPLANATION. The greatest number 

of tens cubed making less than 13824 is 
2 tens or 20. This cube is represented 
by the block marked A, which contains 
8000 little cubical units. To this cube 
is added a layer on three faces, in the 
shape of the blocks B, C, and D. The 
base of each of these three blocks contains 20 x 20 little cubes, and all 
three blocks have 1200 little cubes pressing against the cube A. Now 5824 
is the total number of little blocks that must be put around the three faces 
of cube A. Dividing 5824 by 1200 gives an approximation to the number of 
layers of little blocks. This quotient is 4+. The irregular solid formed by 
the cube A and the three broad blocks B, (7, D must now be changed into 
a perfect cube (last finure) by adding the three oblongs E, F, 6r, having 
20 x 4 little cubes in the lower layer of each, and the little cube H, having 



218 INVOLUTION AND EVOLUTION 

4x4 little cubes in its lower layer. Multiplying 1456, or the total number 
of little cubes in the lower layers of all the blocks added to A, by 4, the 
number of layers found, gives the total number of little cubes added to A to 
make a new perfect cube. Thus, the number of little cubes along one edge 
of the final cube is 24, and 24 is the cube root of 13824. 

Let us compare the several blocks of this solution with the terms of the 
formula 3 + 3 t 2 u + 3 tip -f it 3 . To what block does t 3 correspond ? To 
what set of blocks does 3 fiu correspond ? To what set does 3 tv? correspond ? 
To what block does it? correspond ? 

23. Explain with, the blocks the extraction of the cube root of 
19683, also of 148877. 

SYMMETRY 

144. Symmetry in Algebra. A quantity containing two or 
more letters is symmetrical if these letters can interchange with- 
out altering the value of the quantity. 

Thus, (a -f 6) 2 = a 2 + 2 a?> + 6 2 is symmetrical, because if a and 
6 change places, the quantity has same value as before. Simi- 
larly (a + 5) 3 , (a 4- 6) 4 , (a-ffr-f-c) 2 are all symmetrical. Notice 
a 5 + 5 a*b 4- 10 a s b 2 + 10 a 2 6 3 + 5 atf + b 5 is symmetrical in the 
sense kh&k first part is like last part. 

145. Miscellaneous Exercise in Involution and Evolution. 

1. 2a(-35 2 ?i 3 ) 3 ; a(5a) 3 ; 5(m' 1 )" ; 7(7 p<fi*)* ; a(a m o")'. 

2. VTO; -^/lO; V.~3; V/75; \/37; V^OT; ^\. 

3. (a-6) 11 ; (a- 2 6 +3 c) 3 ; (w 2 + 6 n + 5p - 2 q}\ 

4. ^/(8 c 6 - 60 c 5 -f 114 c 4 + 55 c 3 - 171 c 2 - 135 c - 27). 



, 9^3 

"""" 




6. (2a 2 -7?/) 4 ; VaM^ to three terms ; (1-a + a 2 ) 8 . 



7. -\/-27a 3 a^ 6 ; -v/32 



8. V98; -\/^^5; 



OPERATIONS WITH LITERAL EXPONENTS 219 

OPERATIONS WITH LITERAL EXPONENTS 

146. Simple Operations with Quantities Involving Literal Expo- 
nents. The use of literal exponents was avoided in the earlier 
chapters for the reason that it is difficult for beginners to under- 
stand what they mean. The expression a n means that a, any 
number, is used as a factor n times, n being an indefinite number. 
The same rules apply to literal exponents, of course, as apply to 
specified numerical ones. See 34, 36, 38, 134, 138. 



1. Add 2 x a + 3 y\ 4 x 9 - 6 y\ and 7 x a - 5 #*. ( 29.) 

2. Add 5 ax m , 7 ax m , - 9 ax m , 3 ax m . 

3. From x 2n + y- m take - - 4 y? n 6 y 2m . 

4. Simplify - x 2 " + 1 x a + 2 x 2 * - 5 x 2 ". 

5. Multiply 5 a m by - 2 a\ Ans. 10 a m+n . 

6. 2 x m X x n 7. 3 x m X x. 8. y a X ?/*. 
9. x a X x b x x. 10. 3a-b c x a 2 b e . 11. a"" 1 X a. 

12. a"- 2 xax(-a). 13. x a+l X x a ~ b . 14. 3 a 2 " x - - 5 a 3 ' 1 . 

15. ( 2 a n b m ) (a m b n ). 16. a n+l a n+2 a. 17. 2 a 2 a+1 

18. (a 7l +6 n )(rt n -6 n ). 19. (a n + b n ) 2 . 

20. (a-36 1 ) 2 . 21. (S-3m 3 *) 2 . 

22. (a- m + a; 3 ' 1 ) (a- m 2 or*"). 23. a s -j-a'. 

24. 3.T w+B -f-aj w - n . 25. 

26. (a p+fl + 2b)(a p+q 2 &). 27. 

28. . 3w + ? 3 ' l -r-a; m + ?/ ri . 29. 



Factor the following four exercises : 
30. 0^ + 8. 31. 49x 2m -36y 4n . 

32. 27 a 6 ' 1 - 125 W n . 33. a 2 " + 2a 



CHAPTER IX 

APPLICATIONS OF ALGEBRA 

I. APPLICATIONS IN ARITHMETIC 

147. Percentage. Definitions. Per cent is a Latin word which 
means hundredths. The student should understand the word per 
cent or the sign / to mean hundredths. Percentage may be de- 
nned as analysis in which the fractions denoting the relations 
of the numbers are expressed as hundredths : or as proportion in 
which the ratio (the rate) which enters has 100 for its denominator. 

The base in any problem is the standard number or the 
number that x would ordinarily stand for in algebra. It is the 
number of which the per cent is taken. All the other numbers in 
the problem are described as so many hundredths of the base. 
The percentage in a problem is the number which is a certain num- 
ber of hundredths of the base. The amount is the sum of the base 
and percentage. The difference is the base less the percentage. 

148. Table of Terms and Rates, using n to denote the Number of 
Per Cent. From the definitions in the preceding article, we have 



Base = 100 % of base. 



n 



Percentage = n % of the base (or x base). 

Amount = 100 % of base +n % of base =(100 +ri) % of base. 

Difference = 100% of base n% of base = (100 ri)% of base. 



General Rule for Percentage. Get 1 % (or ^ or -J-, etc.) of the base 
by dividing a, given number in the problem by the number of per cent 
(or thirds, fifths etc.) corresponding to it. From this quotient, find 
the answer sotujht. 

220 



TABLE OF TERMS AND RATES 221 

1. What is 20 % of 175 ? 

SOLUTION. 1 % of 175 = 1.75 ; then 20% of 175 = 20 x 1.75 = 35. 

2. What is 85 % of 1.90 ? 2i % of 160 ? 15 % of 32 ? 

3. A country miller takes for toll 6 qt. for every 5 bu. of 
grain ground. What per cent does he get ? 

SOLUTION. 1 % of 5 bu., or 100 qt. = ] .6 qt. Then 6 qt. equals as many 
per cent of 5 bu. as 1.6 qt. is contained times in 6 qt., or 3.75%. 

4. 16 is what per cent of 64 ? 25 is what per cent of 200 ? 

5. A man cleared $25 by selling a horse at a gain of 121 cj Qt 
How much did the horse cost ? 

SOLUTION. 12 \ % of cost of horse = $ 25. 

1 % of cost of horse = ^ of .$ 25, or $ 2. 
100 % of cost of horse = 100 x $ 2 = $ 200. 

Or, since 12| % = |, \ of cost of horse = $ 25 ; then cost of horse = 8 x 
$25, or $200. 



6. 60 is 40 % of what number ? 15 is 3 % of what number ? 

7. By selling coal at $6 a ton a merchant lost 25 %. What 
per cent would he have gained by selling it at $ 9 a ton ? 



ALG. x T 2 ^ x = 6 ; .-. x = S. 



SOL. 






100 



AIUTH. 75 % of cost = $ 6. 
SOL. 100% of cost = $8 



l%of$8=.08; $l-j-|.Q8=12.5. 

8. What number diminished by 12% of itself equals 44? 
What number increased by 12 % of itself equals 560? 

9. A town whose population increased 20 % during a decade 
had 1500 at last date. What was the population at first ? 

10. An architect charges 2.1 % for plans and specifications, and 
1 % for superintending. How much does he gain if the building 
costs $14,902.50? 

11. A man sold two houses for $3600. On one he gained 25% 
and on the other he lost 25 % . Did he gain or lose, and how much ? 

12. A dealer is obliged to sell sugar so that for 43.5 Ib. he 
receives as much as 36 Ib. cost. What per cent does he lose ? 



222 APPLICATIONS OF ALGEBRA 

149. The Formulas of Percentage. The problems of the preced- 
ing article were given partly to bring again into the student's 
mind the language and reasoning of percentage. We are now in 
position to treat the subject of percentage algebraically. 

Let b = the base, r = the rate, p = the percentage, a = amount, 
and d = difference. Then by the preceding article 

(1) p = br-, (2) a = b (1 + r) ; (3) d = b(l - r). 

1. Solve Eq. (1) for 6j also for r. 

SOLUTION. b = P- (Div. Ax.); r =& (?). 
r b 

2. Translate the results just found into words. 

SOLUTION. From b = P- we get this rule : To find the base when the per- 

r 

centage and rate are given, divide the percentage by the rate. 

Translate r = P- likewise into a rule. 
b 

3. Solve Eq. (2) for b and translate the result into words. 

4. Solve Eq. (3) for b and translate the result into words. 

5. Solve by the formula method. What is 20 % of 175 ? ( 148, 1.) 

SOLUTION. Here we have b and r given to find p. b 175, and r = .20 
By the formula p = br. Then p = 175 x .20 = 35. Ans. 

6. Solve the other exercises of 148 by the formula method. 

150. Interest is money paid for the use of money at a given 
rate per cent per year. Two general methods of counting interest 
may be distinguished : (1) the Aliquot Parts Method ; (2) the 
Cancellation Method. 

The principle of the aliquot parts method is to take a problem 
whose conditions are nearly the same as those of the given prob- 
lem, but which is easy to calculate, and to change the result 
found by using aliquot parts, by easy steps to the required one. 
It depends on the distributive law ( 13, 5). 

We will solve problems by the aliquot parts method and check 
the result by the cancellation method. 



INTEREST 223 

1. What is the interest on $248 at 5 % for 3 yr. 9 mo. 9 da.? 

(1) ALIQUOT PARTS SOLUTION. 

1 year's interest on $ 248 at 5 % =$12.40 

~3~yr. interest on $ 248 at 5 % = 37.20 

6 months' (= \ of 1 yr.) interest 6.20 

3 mo. ( \ of 6 mo.) interest = 3.10 

9 da. ( =: -fa of 3 mo.) interest = .31 

3 yr. 9 mo. 9 da. interest = $ 46.81 

(2) CANCELLATION SOLUTION. 

.62 15.1 

$ 2".$; x 5x^ = $46.81. (Check.) 

r-r 
3 

EXPLANATION OF CANCELLATION METHOD. Here one per cent of the 
principal is taken and multiplied by the given number of per cent giving the 
interest for one year. This product is multiplied by the number of years 
(or number of months divided by 12) the money was loaned; the result is 
the interest required. 

One has a number of choices in getting the time factor. (1) To get the 
time in years and fractions of a year, common or decimal. (2) To get the 
time in months and divide by 12. (3) To get the time in days and divide 
by 360 or 365. To reduce to months and divide by 12 is to follow the middle 
course. 

To reduce the time to months ; multiply the number of years by 12 and add 
the number of months ; annex as a decimal the number of days divided by 3. 

EXPLANATION. 3 days make .1 of a month, and there will be as many 
tenths of a month as 3 is contained in the given number of days. For 5, 10, 
20 days use common fractions. Thus, 10 days = \ month rather than .3| 
month. 

Find the interest in the following problems by both cancellation 
and aliquot parts methods. See that the answers agree. 

2. $ 360 at 5 % for 2 yr. 5 mo. 16 da. 

SUGGESTION. In aliquot parts method get the interest for 1 yr., then for 
2 yr., then for 4 mo., then for 1 mo., then for 15 da., then for 1 da. Add 
results exclusive of first, Evidently 4 mo. = | yr. ; 1 mo. = of 4 mo. ; 

15 da. = I mo. ; 1 da. = T V of 15 da. 

20 
In the cancellation method the statement is 3.6 x 5 X -' ' 




224 APPLICATIONS OF ALGEBRA 

3. $ 318 at 6 % for 1 yr. 7 mo. 19 da. 

4. $525, 7 %, 3 yr. 10 mo. 25 da. 



1.75 

SUGGESTION. ^ X 7 x $^ - ? 

^X 
i 
5. $296.50, 7%, 7 mo. 22 da. 

SOLUTION. When this problem is set down we find there is no chance to 
cancel. So we change the rate to 6 %. To the answer thus obtained we add 
\ of itself to get the interest at 7 %. 

3.8f 6111.4647 

o I _ 



$2.965 x $ x = $11.4646. 



$13.38 Ans. 



The solution in this way is practically the same as the so-called six per 
cent method. 

Whenever no good opportunity offers to shorten the work by cancella- 
tion, the rate should be changed to 6, or 4, or 8, or 9%, and the answer thus 
obtained changed to required rate by the aliquot parts method. 

6. $ 647.20, 5 %, 1 yr. 11 mo. 7 da. 

7. $943.70, 7%, 2 yr. 2 mo. 10 da. 

8. Find the interest on $ 153 at 7 % for 67 days. 

SOLUTION. $ 1.53 x 6 x -^ = $1.53. 

Interest on $ 153 at 6 % for 60 days = 8 1.53 

Interest on $ 153 at 6 % for 5 days (= ^ of 60 days) = .127 

Interest on $ 153 at 6 % for 2 days ( = -fa of 60 days) = .051 

Interest on $ 153 at 6 % for 67 days = $ 1.708 

Interest on $ 153 at 1 % (= | of 6 %) for 67 days = .285 

$1.99 

9. $197, 6%, 64 da. 10. $12.50, 7%, 73 da. 

11. $648, 5%, 76 da. SUGGESTION. $6.48x5x^=36.48. 

SUGGESTION. 

12. $939.72, 8 %, 53 da. $0.3072 x 8 x /^ = $9.3972. 



13. $786,94, 9 %, 98 da. 14. $254.39, 3 %, 127 da. 



THE FORMULAS FOR INTEREST 225 

151. The Formulas for Interest. To treat interest algebraically, 
let p == principal, r = rate, = time in years and fractions of a 
year, i = interest, a = amount. Then by the definition of interest 

( 150) 

(1) prt = /. (2) a=p+prt = p(l + rt). 

The problems of the preceding article were all solved by Eq. (1). 
To show how the aliquot parts method uses this formula and the 
distributive law we will state Ex. 1, 150. 

$248 x .05 x (3 + i + i of i -f T V of i) = $46.81. 



1. Solve Eq. (1) first for p, then for r, and last for t. 

SOLUTION TO IST. { . . 

p = - (Ax. ?). 
rt 

2. Translate the three results just found into words. 

SOLUTION. From p = we get this rule : To find the principal where the 

rt 

rate, time, and interest are given, divide the interest by the product of the 
rate and time. 

Or, by introducing the factor 1 (thus, p=- l \ : Divide the given interest 

by the interest on $ 1 at the given rate for the given time. 

3. What principal will produce $ 29.50 interest in 2 yr. 6 mo. 
at 5 % ? Solve by substitution in the formula. 



4. What rate will produce $26.07 interest from $237 in 2 yr. 
2 mo. 12 da. ? 

5. Solve Eq. (2) above for p. In this case p is called present 
worth) and / is called true discount. 

6. Solve by the formula the problem to find the present worth 
of $200 due in 9 mo. 15 da., allowing 4% as the interest rate. 

7. Change the result in 5 into a rule. 

8. Solve by the rule just obtained the problem of finding the 
present worth of $ 500 due in 1 yr. 6 mo. 20 da., using 6 % as the 
interest rate. 



226 APPLICATIONS OF ALGEBRA 

152. Price and Cost Problems. As a rule, price means the cost 
of a unit quantity of anything and cost the total value, though 
these meanings are often interchanged in common language. 

Thus, what is the cost of 30 Ib. of sugar when the price is 5^ 
a pound ? What is the price of ham when 5 Ib. cost 80 / ? 

Such problems are evidently solved by the simple operations 
of multiplication and division. In the operation of multiplica- 
tion either factor can be distributed, that is, separated into parts. 
(See 13, 5 for distributive law which underlies all operations 
of multiplication.) The aliquot parts method may distribute 
either or both of the two given factors. 

In this article, work by aliquot parts and check by a cancella- 
tion solution. In the aliquot parts solution the student, should 
visualize as much as possible. 

1. What is the cost of 2675 Ib. of coal at $ 5.40 a ton? 

Cost of 2000 Ib. = $5.40 

500 Ib. ( = ton) = 1.35 

100 Ib. (=|of 500 Ib.) .27 

501b. (=iof lOOlb.) .135 

25 Ib. (= i of 50 Ib.) _ = .067 
2675 lb.~ ~^T$7.22 

In this solution the number of pounds is distributed into parts. 

2.70 9fr ,, 
SOLUTION BY CANCELLATION. $.0 x =^^ - $7.22. (Check.") 



2. What is the cost of a farm 70 by 120 rods at $ 93.50 an acre ? 

Cost of 40 x 120 rods, or 30 acres, at $93.50 = $2805.00 

20 x 120 rods = | of preceding = 1402.50 

10 x 120 rods = | of preceding = 701.25 

70 x 120 rods = $4908.75 

In this solution the land is distributed into strips. The problem can also 
be solved by first finding the number of acres, and then getting the cost at $ 90 
an acre, then at $3 an acre, then at $ .50 an acre, and adding these results. 

3. W T hat is the cost of a farm 90 by 140 rd. at $67 an acre ? 
SUGGESTION. Begin by finding the cost of a tract 80 by 140 rd. 



PRICE AND COST PROBLEMS 227 

4. Find the cost of a piece of land 46 by 124 rods at $ 85 an acre. 

5. What is the value of a farm 176 rods wide and 228 rods 
long at $ 150 an acre ? 

6. What is the cost of 6850 Ib. of hay at $8.40 a ton? 

SUGGESTION. First get the cost of 3 tons, then of 500 Ib., then of 250 Ib., 
then of 100 Ib. 

7. W T hat is the cost of 4655 Ib. of hay at $12.50 a ton? 

8. What is the cost of 3375 Ib. of coal at $6.50 a ton? 

9. What is the cost of 8956 Ib. of coal at $3.50 a ton? 

10. What is the value of 63,870 Ib. of steel rails at $28 a ton ? 

11. What is the cost of a field 85 by 95 rods at $ 76.50 an acre ? 

12. What is the cost of 19,960 Ib. of feed at $29.10 a ton ? 

13. What is the cost of 21 Ib. 11 oz. of butter at 26 / a pound? 

SUGGESTION. First get cost of 20 Ib., then of 1 Ib., then of 8 oz., then of 
2 oz. , then of 1 oz. 

14. What is the cost of 12 Ib. 13 oz. of lard at 14^ a pound? 

15. What is the cost of 22 Ib. 7 oz. of butter at 23^ a pound ? 

16. What is the cost of a pile of 4-foot cord wood 13 ft. long 
and 5 ft. high at $ 4 a cord ? 

SUGGESTION. First get the cost of a pile 12 ft. long and 4 ft. high ; then 
of a pile 12 ft. long and 5 ft. high by adding of itself to preceding; then of 
a pile 13 ft. long and 5 ft. high by adding -^ of itself to last result 

17. Find cost of a pile of 4-foot cord wood 26 ft. long and 6 ft. 
high at $ 3 a cord. 

18. Find cost of 5 piles of wood, each 7 ft. long, 31 ft. wide, 
and 5 ft. high at $ 3.75 a cord. 

SUGGESTION. Get cost 4 ft. wide and subtract one eighth of result. 

19. What is the cost of 20 planks 18 ft. long, 14 in. wide, 2 in. 
thick at $22 a thousand? 

SUGGESTION. Find the cost on supposition that the planks are 12 in. 
wide and add of itself to the result, 



228 APPLICATIONS OF ALGEBRA 

20. What will be the cost of 10 planks 16 ft. long, 16 in. wide, 
and 3 in. thick at $18 per M? 

21. What is the cost of 25 joists 20 ft. long, 15 in. wide, and 
4 in. thick at $26 per M? 

22. What is the cost of 8855 bricks at $8.40 per thousand? 
SUGGESTION. Get cost of 8000, 800, 50, and 5 and add results. 

23. The preceding problems can be generalized. Let p price, 
n = number of units or articles, c = cost of whole. Then 

c = np. 

24. Solve c = np, first for p, and then for n } giving axioms. 

25. Change (1) c = np, (2) p = - } (3) n = - into words or 
rules. n P 

SOLUTION TO (1). When the price and number of units are given to find 
the cost, multiply the price by the number of units. 

REMARK. We have in the formula c = np, composed of but three letters 
and an equality sign, a quite remarkable content of meaning. It is one of 
the glories of algebra that it can convey so much meaning with so few marks. 
As we have seen, this one formula, either directly or by implication, includes 
as special cases a very large number of the problems of arithmetic ! 

To show the formula c = np includes such a problem as Ex. 1, p. 226, we 

write 

p n c 

$5.40 x (1 + J + I of i + i of i of i + 1 of i of I of 1) = $7.22 ; 
2.70 



or, 



/T7TK7 

a. In the preceding articles the answers were obtained correct 
to cents. Students should try to avoid using more decimal places 
in solutions than are needed. To this end they should know of 
Oughtred's abridged or contracted multiplication, and abridged 
division, which are given in good arithmetics. 

153. Averages. Asa rule the best of a series of values, all appar- 
ently equally good, is the average, obtained by adding the several 
values and dividing the sum. by the number of values added. 



* AVERAGES 229 

If a, 6, c, are a series of values whose number is n, and x is 
the average value, then 

a) *= a+ + c+ '"- 

1. The weights of a football eleven were respectively 165, 170, 
171, 176, 189, 192, 195, 198, 201, 220, 230 Ib. What was the 
average ? Solve by substituting in the formula. 

2. The weights of an invoice of 7 horses were 1463, 1521, 1284, 
1633, 1379, 1523, 1407 Ib. What was the average weight ? 

3. In a certain town there were 249 houses ; of this number 4 had 
12 T ersons in them, 10 had 10 persons, 20 had 8 persons, 40 had 6 per- 
sons, 70 had 5 persons, 55 had 4 persons, 25 had 3 persons, and 25 
had only 2 persons. What was the average number in each house ? 

4. A man measured an iron rail four times, getting 29 ft. 114- in. 
29 ft. llf in., 29 ft. llf in., 29 ft. H T 9 g. He considered his second 
measurement twice as good as his first ; the third three times as 
good as the first ; and the last five times as good as the first. W r hat 
is the best value of the length of the rail ? 

SOLUTION. In such a problem each measurement is multiplied by its 
"weight," and the sum of these products is divided by the sum of the 
"weights." The "weight" of the first measurement is taken as 1 ; then the 
weight of the second is 2 ; that of the third is 3 ; and that of the fourth is 5. 

However, there is no object in using the 29 ft. 11 in. at all, since it will be 
29 ft. 11 in. all through the solution and 29 ft. 11 in. at the end. Hence, we 
use only the amounts over 29 ft. 11 in. We have then 

(1x1 + 1x2 + 1x3 + ^x5) -s- (1 + 2 + 3 + 5) = .54. 

Hence, the weighted average length of the bar is 29 ft. 11.54 in. 

5. A student has 91 % in geometry, 93 % in Latin, 87 % in 
English, and 81 % in science. What is the average ? 

SOLUTION. Instead of adding the grades as they stand and dividing by 4, 
we write ^ 90 + i) + (90 + 3) + (90 - 3) + (90 - 9)] -- 4. 

In the calculation we would first add the four 90's and then divide by 4, 
thus coming right back to 90. Evidently this part of the calculation is useless. 

I I O O Q 

Dropping the 90's we write - = 2. Hence, answer is 90 2, 

or 88%.. 4 



230 



APPLICATIONS OF ALGEBRA 



Average the following grades in the same way, taking 90, or 80, 
or 70, or 75 as a reference, dependent on where the given grades 
lie. Those above the reference number are positive ; those below, 
negative. 

6. 95%, 88%, 97%, 84%. 7. 65%, 79%, 84%, 81 %. 

8. 68%, 73%, 79%, 85%. 9. 81%, 77%, 89%, 92%, 85%. 
10. 62%, 68%, 79%, 85%. 11. 69%, 83%, 90%, 97%, 92%. 

12. A student recited 5 times a week in Latin, 4 times in 
mathematics, 3 times in German, 2 times in science, and 2 times 
in drawing. He got in Latin 90%, in mathematics 80%, in 
German 78%, in science 95%, in drawing 72%. AVhat was his 
average for the term ? 

SOLUTION. Evidently the number of hours a study is taken a week ought 
to be its " weight." Use 80 as the reference. Multiply each excess or defect 
by its weight. Add the products algebraically and divide by the sum of the 
weights. Ans. 83f%. 

13. Average German 4 times a week 75 % ; geography 3 times, 
82 % ; history 2 times, 80 % ; algebra 4 times, 89 %. 

14. Average biology 3 times a week 85 % ; English history 2 
times, 82% ; trigonometry 4 times, 88% ; rhetoric 5 times, 87%. 



II. APPLICATIONS IN GEOMETRY 

154. Some Theorems of Geometry 
stated algebraically with Exercises. 

1. THEOREM. The sum of the 
three angles of any triangle is equal 
to two right angles, or 180. 

Construct on sheets of paper 
several triangles like ABC in the 
margin. Cut the angles off at the 
dotted lines in the figure and add 
them by placing them together 
with a common vertex and common 




SOME THEOREMS OF GEOMETRY 231 

sides. Cut a right angle off from one corner of the sheet of paper 
and measure the sum. It will be found that the sum of the three 
angles is t\vo right angles, no matter what lengths the sides of 
the triangles have. 

Let a, b, c denote the number of right angles, and d, e, f the 
number of degrees in the three several angles of any triangle. 

Then 

(1) a + 6 + c = 2; (2) (Z + 



Solve first equation for a ; for b ; for c : solve second equation 
for d. 

2. One angle of a triangle is 3 times another angle, and the 

third is 4 times the latter. How many degrees are there in each ? 

> 

3. One angle of a triangle is m times another, and the third 
angle is 4 times the latter. How many degrees in each angle 
are there ? 

4. THEOREM. In any right-angled triangle, the square of the 
number of units of length in the side opposite the right angle (called 
the hypotenuse} is equal to the sum of the squares of the number of 
units of length in the other ttco sides. 

Stated algebraically by denoting the lengths of the 
sides by letters as in the margin, the theorem becomes 

c 2 = a 2 + b 2 . 



a 



To test the truth of this theorem, measure 3 inches 
along one side of the sheet of paper and 4 inches 
along the adjacent side from the right angle. Now 
measure the distance apart of the extremities of the two distances 
measured. It will be found to be 5 inches because 3 2 -f4 2 = 5 2 . 
Again, measure 2 inches along one side and 5 inches along the 
adjacent side and see if the distance is V2- + 5 2 = of + . Meas- 
ure still other distances, making the same test. 

5. Solve c 2 = a 2 + b 2 (1) for c j (2) for a ; (3) for b. 



232 APPLICATIONS OF ALGEBRA 

Solve by substituting in the formula just found the following 
problems dealing with right-angled triangles. 

6. If two sides of a right-angled triangle are 6 and 8 inches 
respectively, what is the length of the longest side ? 

7. Given a = 7, c = 12, to find &; a = 16, 6=22, to find c; 
b = 12, c = 40, to find a. 

8. A ladder leaning against a vertical wall stands on a hori- 
zontal pavement 10 ft. from the wall. If the ladder is 22 ft. in 
length, how high up does it reach on the vertical wall ? 

9. A carpenter makes mortises for a brace 41 ft. out from the 
right angle along a beam and the same distance along a post, 
meeting the beam at right angles. What length must the brace 
have ? 

10. How far is it from one corner of a farm 80 rd. by 120 rd. 

to the opposite corner ? 

11. If a, b, and c are the adjacent edges of 
a rectangular box and d the diagonal from one 
lower corner to the most distant upper corner, 
show that d 2 = a? -f- b 2 -f- c 2 - 

SUGGESTION. Denote by m the diagonal of the lower 
face or base of the box. Then what does m 2 equal ? Consider now the right 
triangle whose base is m and perpendicular one of the vertical edges of the box. 

12. Solve d 2 = a 2 + b 2 + c 2 for (1) d ; (2) a ; (3) 6 ; (4) c. (See 
H7, 1.) 

13. What is the length of the longest straight rod that can be 
placed inside a box 1.4 m. by 1.2 m. by 9 m. ? 

14. A tree standing vertically on level ground is 60 ft. high. 
Upon being broken over in a storm, the upper part reached from 
the top of the trunk to the ground just 30 ft. from the foot of the 
trunk. What was the length of the part broken off ? (See 97.) 

15. A rectangular room is 9 ft. high and is 2 ft. longer than 
wide. If it were 1 ft. longer and 2 ft. wider, it would contain 
270 cu. ft. more. What are the dimensions of the room ? 




SOME THEOREMS OF GEOMETRY 



233 



16. Given the base b and the difference d, between the hypote- 
nuse and perpendicular of a right-angled triangle, to find the 
perpendicular. 

17. If the base of a triangle is 6 and the difference between 
the hypotenuse and perpendicular is 2, what is the perpendicular ? 

18. THEOREM. A line drairn parallel to one 
side of a triangle divides the other two sides into 
parts which are in proportion; also, the corre- 
sponding sides of the two triangles formed 

are proportional. (See 72.) 

Thus, if the line n is parallel to the 
side m of the triangle in the margin, 

a + b __ c + d 
a 



a 




then, 



= ; also 
b d' 



and 



c 



a 



m 
n 



19. THEOREM. If the parts of two sides of a triangle are in pro- 
portion, the line dividing these sides is parallel to the third side. 

The truth of these theorems can be tested by drawing figures 
and making measurements. 

20. Solve - - for a: for b: for c; for d. 

b d 

21. If a = 2 inches, c = 1.75 inches, rZ = 4.5 inches, find b by 
substituting in the formula. 

22. To find the distance across a 
stream by theorem in Ex. 18 above. 

To find the distance AB across a 
stream, measure a, 6, and c as repre- 
sented in the figure, b being parallel 
to x. Then x is found from the pro- 




a f* 

portion - = - by solving for x. 

u X 

23. If = 40 ft., ^ = 
c = 540 ft., what is x ? 



ft., and 



234 



APPLICATIONS OF ALGEBRA 



24. To find the height of a tower by measuring its shadow and 
that of a pole at the same time of day. 

SUGGESTION. Let AB or x in the figure to Ex. 22 denote the height of the 
tower, 6 that of the pole, a the shadow of the pole and c that of the tower. 
Find x. 

Make measurements of the height of different objects in this way. 

25. What is the height of the Washington Monument at Wash- 
ington, D.C., if the monument casts a shadow 90 ft. long at the 
time a flagpole 37 ft. long casts a shadow 6 ft. long ? 

26. THEOREM. TJie circumference of a circle equals TT times its 
diameter, or 2?r times its radius (TT = 3.1416, nearly). 

Divide the circumference by the diameter in various sized 
circles and compare the results. 

27. THEOREM. The area of a circle is equal to one half of the 
circumference times one half of the diameter. 

To show this cut a 
circle into ten equal 
pieces as a pie is cut. 
These pieces can be 
arranged as in the 
figure below, mak- 
ing what resembles a 
parallelogram whose 
base is half the circum- 
ference and altitude 
half the diameter. If 
the circle had been 
cut into 1000 pieces 
instead of ten, one 
could not distin- 
guish by the eye the 
lower figure from a 
rectangle. 





SOME THEOREMS OF GEOMETRY 



235 



28. If r=the radius of a circle and a its area, show by eliminat- 
ing c, from a = rx Jc (Ex. 27) and c=2vr (Ex. 26), that a=^. 

29. If d = the diameter of a circle and a its area, show from 
Ex. 28 by eliminating r that a = -J Trd 2 , or a = .7854 d 2 . 

30. Calculate the area of a circle whose radius is 14 inches, 
(1) by using the theorems of Ex. 27 and 26; (2) by using the 
formula of Ex. 28 ; (3) by the formula of Ex. 29. 

31. Calculate in the shortest way the areas of circles whose 
radii are 5, 4, 6, 24; whose diameters are 4, 11, 35. 

32. If r is the radius of a circle, d its diameter, c its circum- 
ference, and a its area, make a formula to find a in terms of c, 
that is, get a value of a which has no letters except c and TT in it. 

33. Find a from the formula just obtained, when c = 7j 9; 12. 

34. Solve the formula of Ex. 32 for c.. 

35. Find c when a = 25, using the formula just obtained. 

36. Show that the area of a circular ring, whose outer radius 
h R and inner radius is r, is w(/2 2 --r 2 ). 

37. Calculate area when R = 20, r = 10 ; also when R 100, 
r = 25. 

38. The point C 
is the middle of 
the straight line 
ACB(=4r). On 
AC, CB, and AB 
semicircles are con- 
structed with these 
lines as diameters, 
all on the same side 

of AB. If x is the radius of a circle which touches the small 
circles externally and the large circle internally, show that x = f r. 

SUGGESTION. Express FC, EF, and CE, using x and r. What kind of 
figure is EOF? 

REMARK. Such a construction might appear in a large window, or in 
laying out a park, or the like. 




F 



236 



APPLICATIONS OF ALGEBRA 



III. APPLICATIONS WITH SQUARED PAPER 

155. Solution of Simple Practical Problems by Squared Paper. 

The solution depends on 154, 18. 

1. Given the price of an article, to construct a graph from 
which may be obtained the cost corresponding to any given num- 
ber of units (or articles), or the number of units corresponding to 
any given cost. 

SOLUTION. Thus, suppose 1 Ib. of tea costs 75^. To construct a graph 
with these conditions, take sixteen 2-millimeter units on the Y-axis above 
the origin to denote 16 oz. or 1 Ib., and fifteen 2 -millimeter units along the 
JT-axis to the right to represent 75^, thus making one 2-milliineter length 

Y 



Oz. 

30 



25 



20 



15 



10 







2 



P 



23 



50 



75 



1.00 



1.25 



1.50 



X 



represent 5/'. Join the point P located by these coordinates with the ori- 
gin 0. Then the cost of any number of ounces of tea at this price can be 
read off the graph by starting at the given point on the Y-axis, running across 
on a horizontal line to the graph line, and then down on a vertical line to the 
X-axis, remembering that each 2-millimeter length ou the JT-axis is 6^. 



SOLUTION OF PROBLEMS BY SQUARED PAPER 287 

Thus, to find the cost of 23 oz. , start at 23 on the F-axis, pass along the 
horizontal line 23 to the graph, then down vertically to $ 1.08. 

Again, to find the number of ounces corresponding to any given cost, start 
with cost, go up vertically to the graph, and then along a horizontal line to 
the required number of ounces. 

Thus, find how many ounces can be bought for $1.25, and check answer 
by an arithmetical solution. 

2. Construct a similar graph for the cost of butter at 35^ a 
pound. Then find by the graph the cost of 21 oz. ; of 27 oz. ; of 
11 oz. How many ounces can be bought for 50^? for 30^? 
Check in each case by an arithmetical solution. 

a. It was stated at the beginning of this article that these graph solutions 
depend on 154, is, which says a line drawn parallel to one side of a tri- 
angle makes the corresponding sides of the given triangle and the new tri- 
angle formed proportional. Now, if we take the X-axis, the graph line, and 
the 1.50 vertical line in the figure in this article, we have a triangle. The 
$1.50 line is thought of as one side and we know the other vertical lines, as 
the $1.00 vertical line, are parallel to it. Thus, we see that 

any no. of oz. : any other no. of oz. = cost of 1st no. : cost of 2d no. 
Of course, ail such problems can be solved by proportion in arithmetic. 

3. Construct a graph for reading off interest, taking 6 % as the 
rate. 

SUGGESTION. Locate a point, using 60 2-millimeter units for the abscissa 
(representing 360 days), and 6 centimeters representing .$6 interest on $100, 
as the ordinate, and join it with the origin. Then the interest on $100 for 
any number of days can be read off by starting at the given point on the 
X-axis, going vertically to the graph, and then horizontally to the F-axis, 
where the answer is read off. Evidently a 2-millimeter unit on the X-axis 
means 6 days, and a 2-millimeter unit on the F-axis 2 



4. Find the interest by the graph just constructed for 96 days 
at 6 % on S 100. For 36 days. For 50 days. How many days' 
interest is $ 4.60 ? $ 7.20 ? $ 2.80 ? 

5. Construct graphs for other per cents and use them in the 
same way. 

6. Suppose an automobile travels 15 mi. an hour. Construct 
the graph for this motion by using 15 2-inillimeter units as 



238 



APPLICATIONS OF ALGEBRA 



abscissa and 6 centimeters as ordinate of the point determining 
the graph. Find from the graph corresponding intermediate 
values of time and distance. Thus, how long does it take the 
automobile to go 11 mi. ? 19 mi. ? How many miles has it gone 
in 38 inin. ? In 73 min. ? Etc. 

7. Construct a graph for reducing centimeters to inches. 
SUGGESTION. Join origin to point (100, 39.37). 

8. Construct a graph for reading off circumferences from radii, 
or radii from circumferences. 

SUGGESTION. Join the origin to (5, 31.416). 

9. Construct graphs for other similar reductions as they may 
suggest themselves. 

*156. Solution of Problems by Squared Paper. 

1. Two men set out from two towns A and B 30 mi. apart 
at the same time, the one from A walking toward B at the rate of 
4 mi. an hour, and the one from B riding toward A at the rate of 
13 mi. an hour. How far from A will they meet, and in how 
many hours will they meet ? 

SOLUTION. Take a piece of squared paper and select two points A and B 
on the same horizontal line and 30 units apart. Each horizontal unit dis- 
tance denotes 1 mi. In this problem we will take 5 vertical units to denote 
1 hour. The diagonal lines drawn on the figure are graphs. The graph AK 

M 






2 Hours 
IHour 

A 
























7- 




































































/ 




































































/ 




































































/ 






































































/ 


































































-i 


/ 




































































V 


< 



































































/ 








^* 


> 


^* 








































































". 


-> 


* 






T 








































r 


/ 






















^. 




1) 










































/ 


\ 




























>. 


N, 




































/ 




































^ 


"-> 
































/ 










































"^> 


> 
























/ 
















































^ 





















/ 
























































^. 


-> 








































































H 

































































































































































































































































































Miles 



10 



15 



20 



25 



30 



SOLUTION OF PROBLEMS BY SQUARED PAPER 239 

for the man walking is constructed in this way : We count 4 units to the 
right from A for the number of miles the man walks in an hour, and then go 
up 5 units for 1 hr., thus getting the point K. Through A and K the line 
AK is drawn. In the same way we count 13 units to the left from B for 
the number of miles the man from B rides in an hour, and then go up 5 units 
for 1 hr., thus getting point L. Through points B and L the line BL is 
drawn. Where AK and BL meet is the point C. This point <7, so to speak, 
is the graphical answer to the problem. It is 7 units to the right of A, that 
is to say, the men met 7 mi. from A, and it took them If hours to meet. 

The equation for graph AC is y = , and for BL is y = - 

VERIFICATION BY AN EQUATION SOLUTION (see 94). Let x = number 
of units from A to where they meet. Then - = = , whence x 7^ 7 mi. 

This shows that the squared paper solution is not perfectly accurate, but 
gives a very good approximation to the answer. 

2. Two travelers start from two places 60 mi. apart, going 
toward each other, the one by automobile going 18 mi. an hour 
and the other by train going 40 mi. an hour. How long after 
they start will they meet, and how far will their meeting place be 
from the first man's starting point? Verify by algebra. 

3. A man made a trip to the country on his bicycle ; but it 
broke down and he had to walk back. He rode 10 mi. an hour 
and walked 4 mi. an hour, and was gone 3 hr. How many miles 
out did he get ? Solve with graphs and check answer by an 
algebraic solution. 

SUGGESTION. Draw one graph from upwards to the right, using 10 
miles as the rate, and the other downwards and to the right from the 3-hour 
point above as origin, using 4 miles as the rate. Where these graphs meet 
is the answer. 

4. One man walking 3 mi. an hour starts to go from Wash- 
ington to Baltimore. A second walking 5 mi. an hour starts 
4 hr. after the first to make the same trip. How far from 
Washington will the second man have to go to catch the first 
man ? Check with an algebraic solution. 

SUGGESTION. Draw the second graph upward and to the right from the 
4-hour point above as origin, using the 5-hour rate. 



240 APPLICATIONS OF ALGEBRA 

5. Suppose, using the data of 113, 9, that a train starts from 
Pittsburg at 9 A.M., traveling 30 mi. an hour, and another 
leaves Harrisburg at 10 A.M., traveling 35 mi. an hour. Where 
will they meet ? 

SUGGESTION. Put the stations at the proper distances on the X-axis. The 
figure will be like that in Ex. 1, p. 238, except that the second graph will be 
drawn from 1 hr. above B instead of from B, In a general way this is the 
train dispatcher's problem. When a new train is put on or when the schedule 
is changed, he finds where the trains meet by the graphical method. 

6. Using the data of 113, 8, suppose a train leaves Boston 
at 4 P.M. going 40 mi. an hour, and another leaves Fitchburg at 
4.30 P.M. for Boston going 30 mi. an hour. Where will they meet ? 

7. A train leaves Harrisburg, traveling 45 mi. an hour, two 
hours and three quarters after a freight which runs 20 mi. an hour. 
Near what station will the passenger catch the freight ? 

8. Two pipes fill a cistern, the first in 3 hr., the second in 
4 hr. How long will it take both to fill it ? Solve by graph 
method and prove by algebraic solution. 

SUGGESTION. Let any horizontal distance AB (say 10 cm.) represent 1, 
or a cistern full. Then draw graph from A to 3 cm. above B, and one from 
B to 4 cm. above A. Where they cross gives time for both. 

9. It takes A 8 da. to do a job, and B 11 da. How long will 
it take both? Check answer by algebraic solution. 

10. It takes A 6 da. to reap a field, B 8 da., and C 13 da. 
How long will it take all working together ? Check answer. 

SUGGESTION. Get time for A and B as in preceding problem. Then take 
the result for A and B together along with C as in the preceding. 

IV. APPLICATIONS IN PHYSICS 

157. Exercise in solving Some Formulas of Physics. It is ex- 
pected that the teacher will use no time in explaining the mean- 
ing of these formulas. The names which some of the letters 
represent are given to show the student that the formulas have 



APPLICATIONS IN PHYSICS 241 

a meaning, so that he may not be repelled by their unfamiliarity. 
The formulas are given also partly to show that algebra has value 
in other branches. Begin by reviewing 118. 

p 

1. Given (7 = , to solve first for R, and then for E. 

R 

REMARK. To solve for any letter means taking that letter to represent 
the unknown number, or the , in the solution. The formula in Ex. 1 ex- 
presses Ohm's law in electricity. 

SOLUTION. CR - E (Mult. Ax.) ; R = , Ans. (Div. Ax.) ; E = CR, Ans. 

O 

Tfi 

2. Solve C = - for E. R, and r in turn. 

R + r 

SOLUTION. Ans. CR + Cr = E (Mult. Ax.) ; CR - E- Cr (Sub. Ax.) ; 

- Cr . _ E- 



R = 



C 



3. Solve - - = - + - f or R, r, and r' in turn. 

R r r' 

SOLUTION TO IST. rr = Rr' + Rr (Mult. Ax.) ; Rr' + Rr = 

R(r' + r) = rr' ; R = -^ (Div. Ax.). 

r' + r 

4. Solve s - 7 (formula for specific gravity) for w and I. 

i 

on V X P ll r, 

5. Solve - = - for x. 

V - x + v P 

SUGGESTION. Clear, transpose unknown numbers to left, known numbers 
to right. Factor and divide by coefficient of x. 

6. Solve E- = - (a formula in electrical work) for h. 

t 

PV P'V 

7. Solve : - = - for each letter in turn (Combination of 

Boyle's and Charles's law). 

-pi 

8. Solve J - (a formula for one electrical cell) for R. 

B + g + R 

9. Solve = - (a formula for lenses) for /. 



242 APPLICATIONS OF ALGEBRA 



10. Solve a = for v and t respectively. 



11 Solve C= -- for E, R, r, and N (Ohm's law in series). 

J\i j~ _i.V / 

E 

12. Solve C ' = - for E, R, r, or N (Ohm's law in parallel). 



13. Solve = for R, h. and r in turn. 

R 2 7TT 

14. Solve Pd = Rs + i for each letter in it. This formula ex- 
presses a law for all machines. 

15. Solve Rs = i mv 2 for R, s } and m in turn. 

16. Solve g : G : : - - : - for g. What is the value of 

A~ (R -f- 



g when h 2 (but not Ji) is put equal to zero ? ( 60.) 



4. 

17. Solve e = - for P. Also for 6, and for m. 

bh s m 

18. Solve g=2x + ?^-^xf or #. 

s 

V. FORMULAS 

158. General Exercise in framing and using Formulas. We have 
seen derived by induction the formulas for percentage, interest, 
cost of merchandise, etc. In this exercise we will study formulas 
in still other subjects. Consult the dictionary freely for the 
meaning of technical words used. 

1. It is taught in arithmetic that the area of a rectangle is 
equal to the product of its base by its altitude. If b is the 
number of units in the length, h the number in the altitude, and 
a the number of square units in the area, make a formula for the 
area. Ans. a = bh. 



FORMULAS 243 

2. What is the area of a rectangle whose base is 20 ft. and 
altitude is 12 ft. ? Solve by substituting in the formula. 

3. Make a formula for the area of a triangle if its area is 
equal to half the product of its base and altitude. 

4. What: is the area of a triangle if b = 16, and h . = 10 ? 

5. Make a formula for the volume of a box (or right paral- 
lelepiped) if I is its length, b is its breadth, h is its height, and v 
is the number of units of volume in it. 

6. What is the volume of a box if 1 = 12 inches, 6 = 8 inches, 
h = 6 inches ? 

7. Show if s = the total surface of a box that s = 2 (Ib -\-lh-\-bh). 

8. What is the total surface of the box described in Ex. 6 ? 

9. Make a formula for the area of a trapezoid if its area, a, 
equals half the sum of its parallel sides, b and b' } times its 
altitude h. 

10. What is the area of a trapezoid whose & = 15, &'=12, h = 7? 
Draw a figure and mark the letters and dimensions on it. 

11. Make a formula for the sum of 
the interior angles A, B, C, etc., of a 
convex polygon. 

SOLUTION. Join the vertices A, B, (7, etc., 
to any point O within the polygon. Then 
each of the triangles formed, AOB, BOC, 
COD, etc., contains 2 rt. angles (by 154, i). 
Then, if the polygon has n sides, all the 
angles within all the triangles will amount to 

TTI T^ 

2 n right angles. Now the angles at which 

are evidently equal to 4 right angles are not 

to be included in this sum, since all we want is sum of ^4, B, (7, etc. Hence, 

if s = number of right angles in the sum of the interior angles of any polygon, 

s = 2 n 4. 




244 APPLICATIONS OF ALGEBRA 

12. Find by substituting in the formulas just obtained the sum 
of the interior angles of (1) a quadrilateral or 4-sided figure 
(ti = 4) ; (2) of a hexagon or 6-sided polygon ; (3) of a decagon or 
10-sided figure ; (4) of a 13-sided figure. 

13. How many degrees are there in one angle of a pentagon, 
or 5-sided figure, all of whose angles are equal ? In one angle of 
a heptagon, or 7-sided figure, all of whose angles are equal ? 

14. The volume of a pyramid equals ^ the product of the area 
of its base and its altitude. Make a formula for the volume 
when base is a rectangle, letting v = volume, I = length of base, 
w = width of base, and h = altitude or height. 

15. Make a formula for the distance s a body travels, if v is its 
velocity, or rate, and t is its time. 

16. Make a formula for the weight w of a body if the weight w 
is equal to the product of its volume v and density d. (Density 
= weight of a unit of volume.) 

17. Fahrenheit-Centigrade Formula. In the Fahrenheit ther- 
mometer, in common use, the freezing point is 32 above the 
zero of the scale, and the boiling point is 180 above the freezing 
point. The Centigrade thermometer, used mostly in scientific 
laboratories, has freezing for its zero point on the scale and 100 
for its boiling point. If /denotes the number of degrees in any 
reading of the Fahrenheit thermometer, and c the number of the 
Centigrade at the same time and place, let it be required to ex- 
press/in terms of c. 

SOLUTION. 100 Centigrade degrees = 180 Fahrenheit degrees. 

1 Centigrade degree = ^- or f Fahrenheit degrees, 
c Centigrade degrees = c x f Fahrenheit degrees. 

But as/ is counted from 32 below freezing, 

/=fc + 82. 

18. Solve the equation just found for c. 

19. Using the formula, find / when c = 50 ; 4 ; 200 ; - 10. 



FORMULAS 245 

20. Using the formula, find c when/=60; 98; 400; -32; 0. 

21. Taking a whole sheet of graph paper, construct the graph 
of/ = -|c-f 32. Let /denote X distances, and c, Y distances. 

22. Use the graph diagram as in 155, 1, to check the results 
of the calculations in Ex. 19 and 20 above. 

23. Use the diagram to reduce from Fahrenheit to Centigrade, 
being gi ven /= 35; -16; 48; 70; 100; 90. 

159. General Exercise in finding Numerical Values from Formulas. 

1. The formula for the surface of a sphere is s = 4 irr 2 . Using 
3.1416 for TT, find the surface of a sphere whose radius r is 6 
inches. 

2. The volume of a sphere is | Ti-r 3 . Find the volume of a 
sphere whose radius r is 12 feet. 

3. The volume of a sphere is also 1 ircfi. Find the volume of 
a sphere whose diameter d is 2 meters. 

4. The volume of a pyramid having an altitude a and a 

rectangle for base whose length is / and width w is Find 

o 

the volume of a pyramid in which a = 16 ft., I = 12 ft., and 

W = 4:.S ft. 

5. Using the formula of 158, 17, find c when /= 20; change 
25, 38, 95, - 15, 219, - 273 from each scale to the other. 

6. A body falling from a state of rest in t seconds moves 
through a distance s, where s = ^ gt 2 , provided the resistance of 
the air is supposed to not retard the body. Find the distance an 
object, as a stone, moves through in 3 seconds; in 5 seconds; 
in 10 seconds. Use g = 32.16. 

7. A stone dropped from the top of a building reaches the 
ground in 3J seconds. How high is the building ? 

8. Find the time in which $324 will draw $64.80 interest at 
5% interest, using the formula prt = i. 



246 APPLICATIONS OF ALGEBRA 

9. A man sold a horse for $161.57, on which he gained 7 
What was the cost of the horse ? Use formula a p (1 + r). 

10. Find p from the formula a 2 = b 2 + c 2 2 pb when a = 7, 
6 = 6, c = 5. 

160. Exercise in translating Formulas into Rules. 

1. Translate formulas in Ex. 1-9 of preceding article into rules. 
Solution of 1st. The surface of a sphere is equal to four TT 

times the square of the radius. 

2. Translate the formulas in 154, 4, 11, 28, 29, into words. 

3. If V is the volume of a cylinder, V= ?rr% where r is the 
radius of its circular base and I is its length. Translate this for- 
mula into a rule. 

4. If V is the volume of a cone, V= \ -rr^h, in which r is the 
radius of its base and h is its height or altitude. Translate this 
formula into a rule. 

5. If Fis the volume of a hollow sphere, V= f 7r(R 3 r 3 ), in 
which R is the radius of the sphere and r the radius of the hollow 
sphere. Translate the formula into a rule. 

*VI. APPLICATIONS IN MANUAL TRAINING AND 

DOMESTIC SCIENCE 

161. Problems in Manual Training, Domestic Economy, etc. 

1. A boy in manual training making a bookshelf finds the dis- 
tance from the top of the lowest shelf to the under side of the top 
shelf is 4 ft. 6 in. He desires to put between these four other 
shelves of inch boards in such a way that the book space will di- 
minish one inch for each shelf from the bottom to the top. What 
will be the several spaces between the shelves ? 

SUGGESTION. Let x = no. of inches between two lowest shelves. 
Then x + 1 = no. of inches to top of second shelf ; 

x -f 1 + (x 1) + 1 = no. of inches to top of third shelf ; etc. 



MANUAL TRAINING AND DOMESTIC SCIENCE 247 

2. The height of a writing desk from the floor is 29| inches. 
Four drawers one above the other are to be inserted, the lower 
three being of the same height and the topmost drawer one inch 
lower. The drawers are separated by three strips of J-inch stuff, 
and the distance to the floor from the under side of the lowest 
drawer is one inch less than the height of the drawer ; also the 
distance to the top from the upper part of the upper drawer is 
one inch less than the height of the upper drawer. Required the 
several heights described. 

3. In a bookstand whose shelves are adjustable, the distance 
from the top of the lowest shelf to the under side of the top shelf 
is 4 ft. 1^ in. If four shelves are inserted each f in. thick, and 
if the second and third spaces from the bottom are each 11 in. 
narrower than the one below it, and the two top spaces are each 
one inch narrower than the one next below it, how far apart shall 
the several shelves be put ? 

4. A farmer noticed that when he had 100 Ib. of milk and 20 
Ib. of cream he got out of both 12 Ib. of butter. At another time 
when he had 150 Ib. of milk and 25 Ib. of cream he got 16 Ib. of 
butter. What part of a pound of his milk is butter, and what 
part of a pound of his cream is butter? 

5. How many gallons each of cream containing 33 % fat and 
milk containing 6 % fat shall be mixed so as to produce 10 gal- 
lons of cream containing 25% fat? 

6. How much 95 % alcohol, 5 % alcohol, and water shall be 
mixed to make 15 gallons of 25 % alcohol, if the number of gal- 
lons of water used is one half of the total number of gallons of 
alcohol used? 

7. The following formula has been obtained by experiments. 
If b = number pounds of butter and /= number pounds of fat in 
100 Ib. of milk, b = 1.16 (/- 0.2). Find by formula number of 
pounds of butter in 70 Ib. of milk testing 4.4 % ; also, find per 
cent of butter fat when 500 Ib. milk yields 19.7 Ib. butter. 



248 



APPLICATIONS OF ALGEBRA 



The statistics in the following table were taken from Bulletin 
28 of the United States Department of Agriculture. It should 
be stated here that the beef per cents given are for porterhouse 
steak. The cuts from the different parts of the animal contain, 
of course, varying per cents. 



FOOD 


PEK CENT OF 
PROTEID 


PER CENT OF 
FAT 


PER CENT OF 
CARBOHYDRATE 


-Ot-'t/I. * 


22 


20 





Bread ...... 


9 


1 


53 


Butter 


1 


85 





Cabbage 


2 


.3 


6 


Corn bread 


7 


5 


46 


Eggs 


15 


10 





Fish (bass) 


19 


3 





Milk 


3 


4 


5 


Oatmeal ...... 


1(3 


7 


68 


Peas (green) 


7 


1 


17 


Potatoes. 


2 


.1 


18 


Rice . ... 


8 





79 


Wheat 


14 


2 


72 



8. A normal daily ration as given by one authority consists of 
4.5 oz. proteid, 2 oz. fat, and 18 oz. carbohydrate. How many 
ounces each of bread and butter are needed to contain the right 
amounts of proteid and fat? 

SOLUTION. Let x = no. oz. of bread, y = no. oz. of butter. 
Then, using the per cents given in the table, we have 

(1) .09 a; + .Oly = 4.5 

(2) .Olar+ .85^ = 2 
(2i) . 09 a; + 7.65 y ==18 

7.64y = 13. 
y = 1.8 
x = 50 



(Mult. Ax.) 
(Sub. Ax.) 

(approximately) Ans. 
(approximately) Ans. 



NOTE. This result shows that bread itself approximates to a normal 
ration, since only a little butter needs to be added. 



PROBLEMS IN MANUAL TRAINING 249 

9. To contain 4.5 oz. of proteid and 2 oz. of fat, how many 
ounces each are needed (1) of bread and beef ? (2) of bread and 
eggs? (3) of bread and milk? 

10. To contain 4.5 oz. of proteid and 2 oz. of fat, how many 
ounces each are needed (1) of beef and eggs? (2) of bread and 
fish? (3) of oatmeal and milk? 

a. Notice in these problems that an excessively large number of ounces 
of any food would be out of the question, and that a negative answer denotes 
impossibility of fulfilling the requirements. 

11. To contain a daily ration of 3.1 oz. of proteid food and 
18 oz. of carbohydrates, how many ounces each are needed (1) of 
rice and milk? (2) of bread and cabbage? 

12. Dinner for an English canal laborer consists of bread, beef, 
and tea. If the food contains 2.48 oz. of proteid matter and 
1.68 oz. fat, how many ounces of each are used? 

13. If an agricultural laborer in Ireland gets .68 oz. proteid 
food, and 5 oz. carbohydrate from a supper of potatoes and corn 
bread, how many ounces of each does he eat? 

14. The daily ration in the German army in war contains 
4.7 oz. of proteid, 2 oz. fat, and 17.5 oz. carbohydrate. How many 
ounces each are needed of bread, peas (with bacon added to make 
fat 5 %), and eggs to make up this ration? 

15. If a daily ration for a girl of nine years is desired of cereal 
(ground wheat), milk, and eggs consisting of 2.44 oz. proteid, 
1.62 oz. fat, and 7.21 oz. carbohydrate, how many ounces of each 
ingredient are needed? 

16. If Portland cement contains 25 % of silica, 10 % of alumin- 
ium, and 65 % of lime, and limestone rock containing 60 % of lime, 
24% of silica, and 4% aluminium, chalk containing 70% lime, 
9% silica, 2% aluminium, and clay containing 3% lime, 50% 
silica, and 45% aluminium, are mixed to make the cement, how 
many cubic feet of each are needed to make 10 cu. ft. of cement ? 



CHAPTER X 

FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

I. FRACTIONAL EXPONENT EXPRESSIONS 

162. Formulas giving the Laws of Exponents in the Fundamental 
Operations. The laws for integral exponents were given in 34, 
36, and 38. They can be expressed in a few formulas. 

1. Addition and Subtraction Formulas : 

ax m + bx m = (a + b)x m -, ax m - bx m = (a - b) x m . 

2. Multiplication Formula : x m x x n = x m+n . 

yin 

3. Division Formulas : x m -f- x n = x m ~ n j - = x m ~ n . 

x n 

4. Power Formulas : (x m ) n = x mn ; (xyz) n = x n y n z n . 

m m p 

5. Hoot Formulas : -tyx = x n ; \J x m y* x"y\ 

Evidently m in the fractional exponents of the last line denotes 
the power to which its quantity is to be raised, and is called the 
exponent or index of the power; and n denotes the root to be 
taken, and is called the index of the root. 

6. It may be proven that any letter in formulas 1-5 may have 
any value except when a divisor becomes zero. Thus, any 
letter in the preceding formulas may have any value, positive, 
negative, integral, fractional, or irrational, as V3. There are 
limitations, however, that have to be put on the meaning in cer- 
tain cases, as will be explained later. 

250 






FRACTIONAL EXPONENT EXPRESSIONS 251 

Simplify the following, noting which formula applies, and how. 

7. 3x 7 + 4x 7 =? 8. x*xx 5 =? 9. 2z 4 -- 2 . 

10. 9x 3 x4ar>. 11. 12 a 4 -=-3 a. 12. 5m 2 2m 2 . 

13. (ra 2 ) 3 . 14. ^a 5 . 15. 7a^-2a 

16. 3a 3 x7a 2 x2. 17. a 4 x 3 a,- 3 -=- a 2 . 18. S^ 12 -h (a 2 ) 2 . 



19. V^y. 20. (a& 2 ) 3 . 21. 24a 3 6 2 -6a6. 

163. Rules giving the Laws of Exponents, whether Integral, Frac- 
tional, or Negative, in the Fundamental Operations. 

1. In addition add the coefficients of similar quantities for the 
coefficient of the common part in the sum. 

2. In multiplication add the exponents of the same quantity in the 
factors for the exponent of this quantity in the product. 

3. In division subtract the exponent of a factor in the divisor 
(denominator) from the exponent of the same factor in the dividend 
(numerator) for the exponent of this quantity in the quotient. 

4. In raising to powers multiply the exponent of a factor quantity 
by the index of the power to which it is to be raised for the exponent 
of this quantity in the answer. 

5. In extracting roots divide the exponent of a factor quantity by 
the index of the root for the exponent of this quantity in the answer. 

The formulas 162, 1-5, and their translation into words, 163, 
1-5, should be thoroughly memorized. 

164. Exercise in dealing with Quantities containing Fractional Ex- 
ponents. 

1. We have Vc? = a 2 ; -\/o~ 6 = a 2 ; v / a i2 = a 3 . Then, if we fol- 
low the same rule in solving vV, we get -\/a 7 = d s . Similarly, 
Va? = cA Thus, as stated in 162, 5, the numerator of a frac- 
tional exponent denotes the power to which its quantity is to be 
raised, and the denominator denotes the root to be extracted. 



252 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

2. Calculate 8*. 

SOLUTION, v^ = 2 ; 2 2 = 4. Thus, 8' = 4. 

It shortens the labor in such problems to get the root first and 
the power required afterwards. 

3. Calculate in the same way : 4^ ; 16- ; 27^ ; 64^ ; 8* ; 4^ ; 
16'; 125*; (i)*5 (^) J (-27)* 5 (-216)*; 1*; 0*. 



4. 

SUGGESTION. What does the exponent \ denote ? ( 162, 5.) 



5. (4a 2 -12aZ> + 96 2 ) = ? (a 3 - 3a 2 6 + 

6. Express with fractional exponents : Vaj 6 (-4ns. 



/ 

7. Express with radical signs m (^4?is. v^i 3 ); c 3 ; r 3 ; ?/ 

71 

. HI _ 3 

s. 5Vm 3 ); 2 a 5 ; 3a^j 5?/ 4 ; aa;*. 



8. Simplify 2 a^ x 3 a^. Ans. 6 a 5 . 

EXPLANATION. The coefficients 2 and 3 are multiplied together, and the 
exponents \ and | are added for the exponent a in the product, just as in the 
multiplication of quantities with integral exponents. 



9. 2ox5o 10. 3ax4c 11, 3ax-- 

12. 2 m* X 5 mi 13. aM x a-fri 14. M x a 

15. ^-j-aji 16. -4a^-j--a'. 17. 9 

18. m i x ^- 19- (w*). 20. (3 a 1 ) 3 . 

wi* (^Ins. 27 a*). 

21. (5a ! ) 5 . 22. (a*)*. 23. 



FRACTIONAL EXPONENT EXPRESSIONS 2.38 

24. Multiply x + x*+2 by x-\-x*--2 and check product by 
letting x = 9. The 9 is chosen so that its square root can be 
extracted. 

SOLUTION. x + x* + 2 14 

- = x 10 

140 



, -f a; + 2 



33. Divide a 3 6 a 3 b 3 16 b 3 by 

SOLUTION. as 6 as&s - - 16 &5 

at + 2 



- 8 



_ 8 a - 16 



Prove by multiplying divisor and quotient together. 



,4 _ 2,1 

34. 



x 2 + 2 2 + x - 4 = 140 

25. (y* + y* -f 2/*-f-l)Q/^ -1). 26. 

27. (a 3 -f a"3& + b 3 )(a 3 - - a 3 b 3 -f 6^. 28. 

29. (x^y* + a' 5 ?/ 3 + y 7 ) x 3 y 3 . 30. (.1^ -f- 5) (V s - 4). 

31. (a-\-a 2 b^a 3 b T )(a-\-a^b r2 -\-a^b 3 ^). 32. (a 3 -- 4)(a 5 + 3). 



35. (8 m + 12 w 3 + 6 mn^ + ?l 2 ) H- (2 m + 

36. (a 2 - 3 a + 6 a* - 7 a + 6 a* - 5 a^ + 1) -r- (a - a^ + 1). 

37. (x* - xy* + x^y -- y*-) (x + x%y% + ?/). 



254 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

165. Quantities involving Zero and Negative Exponents. 

a 5 1 

1. We have - - = by dividing both terms by a 5 ( 60) ; 
a 7 or 

i a 5 _ _ 2 (By subtracting the exponent in the divisor from 
aiso - el . 

a 7 the exponent of the same letter in the dividend 

for the exponent of that letter in the quotient.) 

a~ 2 = ' (Things equal to the same thing are equal to 

o 

each other, by Axiom 7.) 

It is clear that by taking other exponents than 5 and 7 the 
negative exponent could be made any number. 

Hence, any quantity ivith a negative exponent is equal to one 
divided by the quantity with the sign of its exponent changed. 

ab~ z 1 1 

In ;-, if we replace b~ 3 by and c~ 2 by , we get 



or. (70.) Hence, -=. 
T~ Wd c~-d b s d 

- 2 Xd 
cr 

Tims, we see that any factor of the numerator or denominator can 
be transferred from the numerator to the denominator or from the 
denominator to the numerator of a fraction by changing the sign of 
its exponent. 

Integral quantities having negative exponents are supposed to 
have 1 for denominator. 

Change the following expressions to equivalent ones with posi- 
tive exponents, simplifying when possible : 



. Ans. 3. T^T- 4. a~ 3 . 

ab 2 a~-b- 



6. .. .. 7. 



8. o^xcr 4 . 9. a 2 -=-a- 2 . 10. (a 3 )- 4 . 11. (m~ 2 ) 






ZEKO AND NEGATIVE EXPONENTS 255 

3 b 2 c 

12. To study zero exponents take the problem 

3c 

Dividing both terms by b' 2 , we get -- ( 60.) 

Cv 

But if we follow the rule of division and subtract the exponent 
of b in the denominator from that in the numerator for the ex- 

3 bc 
ponent of b in the quotient, we get - -- Thus, 6 means that b is 

\AJ 

used no times as a factor of the result, that is to say b has dropped 
out of the result. 

b 2 b 2 

We have - = 1 ; also - - = 6 ; therefore b = 1. Or. 
tr cr 

^4?i?/ quantity whose exponent is zero is equal to unity. (See p. 274.) 

13. What is the value of a n ~ 2 when n 2 ? 



14. Simplify ; . 15. 1000' = 

2 



2 r> 12 

16. 3 3 =? 3 2 =? 3 J = ? 3 = ? 3-*=? 3" 2 = ? 3~ 3 

Express with positive exponents : 

17. or 4 /. Ans. %- 18. a;- 1 ?/- 2 . 19. m~%*. 

a; 4 

20. Sm- 1 ^. 21. 4cf*. 22. a-^V 3 . 23. 8 a~ V. 

Write the following fractions without denominators : 
24. -^-- ^%s. 3a-W. 25. -4-- 26. 



(* ** n -/i- 1 ns* ** 

y C(/ L/ *v 

2 3 1 Q 1 

27. -^-- 28. ? - 29. ^ - 30. 



a* 



166. Miscellaneous Exercise in the Use of Fractional and Negative 
Exponents. It is best to begin by changing any quantities written 
in the radical sign notation to the fractional exponent notation; also, 
as a rule, to change negative exponents to positive ones as ex- 
plained in 165. However, in many exercises it is shorter to 



256 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

follow the simple rules for exponents, 163. Thus, in such an 
expression as (2cT*)~ 3 9 it is better to follow the rule of multiply- 
ing the exponent of each factor by the index 3 of the power. 

1. 25*. 2. 4- 2 . 3. 16~i 4. a 3 x a~ 2 . 



2. 4- 2 . 


3. 16~*. 


4. 


6O"~T 
o . 


7. (-8)*. 


8. 


10. (a*) 2 . 


11. (a- 4 ) 2 . 


12. 


14. ( a^) 4 . 


i <=> /'S"^ 4 

J. O. I O 1 


16. 


18. (x*)^. 


1Q -\ / 7^ _s_ -\ //r 

J.J7. V w ^*^ \ " 


20. 


22. (8i~M. 


9^ ^TM-^'tN-S 

A .) ' //(/ IV I m 


24. 









5. a 2 -*- a~ 4 . 6. 8~i 7. ( S)i 8. x* + x 

9. x% -f- af i 10. a* 2 . 11. a~ 42 . 12. aM x 

13. 

17. (A/3) 8 . 18. (a;*)*. 19. #?-*- V. 20. ^2 X \/2~ 3 

21. 2-5-V2. 



Solve the following fourteen problems mentally) quoting in each 
case the theorem used. See 45- 47. 

25. (a* + &*) (a* b^). 26. (a 6)-s-(a* -- 6^). 

27. (a^-6*) 2 . 28. (a + 1 - cr 1 ) 2 . (137, 14.) 

29. (a;* - 6) (a* + 5). 30. (or" - 3 Jr 2 ") 3 . 

31. (^ + 2) 3 . 32. 

33. (x + 27)-5-(x3+3). 34. (16 - 4 -24a- 2 

35. Factor 9 or 2 . 36. Factor 27 - - b~ 3 . 

37. Factor a 2 + 2 + a~ 2 . 38. Factor 12 -- a; l - x~ z . 

39. 

40. (4 a 2 - 9 + 30 a- 1 - 25 a~ 2 ) -5- (2 + 3 a' 1 - 5 a~ 2 ). 

41. (3a- 5 -4a- 4 +5- 3 )(2a- 2 -a- 1 ). 

42. y(9a-12rt* + io_4a^ + - 1 )- (140.) 

43. J/(x% - 6 oj + 15 ^ - 20 + 15 af ^ - 6 a- 1 + a;-*). 

44. (a?* - 2 7/3-)* ( 136) ; (2 ar 1 + 2/~^) 4 ; (3 af * -- 5 ?/- 2 ) 3 . 






RADICAL EXPRESSIONS 257 

II. RADICAL EXPRESSIONS 

167. Radical Quantities. An indicated root of a quantity is 
called a radical quantity, or simply a radical. Thus, V4, A/7, 5-. 






In 3 V2a 3 , or 3 (2 a 5 )"*, 3 is called the coefficient of the radical, 4 is 
the index of the root, and 2 a 5 the radicand. 

Radical quantities may be either rational, as V}, or irrational, 
as V5. A rational quantity is the ratio of two integral numbers. 
An irrational (or surd) quantity cannot be the ratio of two inte- 
gral numbers. To show this by illustration, we note 

I = -625, 
[ f = .428571428571428571 ., 

while V5 = 2.2360679775 .... 

Notice that values of surds neither end nor repeat figures in sets. 

168. Principles used for Reductions in Radicals. We have seen 
( 163) that the same rules hold for fractional as for integral 
exponents, 

; If, then, (abc) 2 = aW, (a&c) 8 = aW, etc., 

it ought to be true that 

(abc)% = aMc^, (a&c)* = a*6*c*, etc. 
' Thus, (4 x 9)* = 4* x 9^ ; (8 x 27)* = 8' x 27* 

; We have (4 x 9)* = 6, and 4^ x 9 * = 6. 

If, now, the sign on one side of (4 x 9)* = $ X 9^ were taken -K 
and that on the other side , the equation would not be true. 
To avoid this difficulty, only positive roots are used. 



C hanging formula, (abc---) 11 = a n b n c u into a ruUj we have 

1. Fundamental Principle. A root of a product is equal to the 
product of the same root of the several factors ; and conversely. 



258 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

By means of this principle any factor within a sign denoting a 
root may be removed outside the sign provided the desired root 
of the factor can be extracted. 

Thus, 12* = (4 x 3)* = 4* x 3* = 2(3)*, 

16* = (8x2)* = 8^x2* = 2(2)* 

2. If both terms of a fractional exponent are multiplied by the 
same number, the value of the quantity is not changed. 

Here, again, we find it necessary to limit ourselves to arith- 

1 2 

metical numbers. For, while 4 T = 4, 4* may equal either 4. 

3. A quantity with a fractional exponent may have its value cal- 
culated either by raising to the power first ( 162, 5) and extracting 
the root afterwards, or vice versa. 

Thus, ' 8* = (8 2 )* = (8*) 2 = 4. 

This principle also holds true only for arithmetical numbers. 
Thus, 4* = 16* = 4, while (4^) 2 = + 4 only. 

169. The Double Notation for Radicals. We have already seen 
( 164) that fractional exponents as well as radical signs denote 
roots. The radical sign to denote roots came into use (1551 A.D.) 
a century before fractional exponents were thought of. Unfor- 
tunately, the existence of this duplicate notation has increased 
for students the difficulty of learning the subject of radicals. 

Change the following quantities from the radical sign to the 
fractional exponent notation, or vice versa : 

1. -\/S. Ans. 8*. 2. V162. 3. 

4. </2c?. Ans. (2 a 8 )*. 5. V5^&. 6. 

7. 6V7. Ans. 6(7)*. 8. 7^/64*. 9. 2a(3a 2 )*. 



10. (i)~*. Ans. v'tt)- 1 . 11. m[ - ) 12. 

Uy 






SIMPLIFICATION OF RADICAL QUANTITIES 259 

170. Simplification of Radical Quantities. A radical quantity is 
not in its simplest form : 

1. If some required root of it can be extracted. 
Type form : (a m )= a\ 

(49 a 2 )* = 7 a ; (27 6 3 )* = ((3 6) 3 )* = (3 6)1 

When possible, express the radicand as a perfect power of a quantity 
ivhose exponent is a factor of the index of the required root ; then 
divide this exponent by the index of the root for the exponent of the 
quantity in the answer. 

2. If any factor of the radicand can have the required root taken 
of it. 

Type form : (a n b) n = a(b) n . 

5(12)* = 5 (4x3)* = 10(3)* ( 168, i) ; (32 atf = 2 a(4a)i 

If the required root of any factor of the radicand can be taken, 
that factor may be removed from the radicand and its root used as a 
factor of the coefficient of the radical. 

3. If the radicand is fractional. 

i 

/aV 1 - 

Type form : - = - ( a b n ~ l ) n . 



= (f x I) ( 60) = ( T V x 21) = |(21) (s 



o o 



(168,1). 



To simplify a quantity whose radicand is fractional, multiply both 
terms of the radicand by a quantity that will make the denominator 



260 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

a perfect power of which the required root can be taken. Extract 
this root and write the result as a factor of the denominator of the 
coefficient. If possible, simplify the numerator as in 2 above. 

4. If there is an irrational quantity in its denominator. This 
kind of simplification will be taken up later ( 177). 

171. Exercise in Simplifying Radical Quantities. All problems 
given in the radical sign notation should be immediately changed 
into the fractional exponent notation, then worked, and last of all 
changed back. Before changing an answer back into the radical 
sign notation, quantities with improper fractions as exponents should 
be separated into two factors, one of which has an integral exponen* 
and the other & proper fraction exponent. 

Thus, 3 a* = 3 a (a)*, which becomes 3 a -\fa?. 

Check answers by extracting roots, as in the last chapter, when 
ever uncertain, assigning numerical values to the letters. 

1. (160*)*. Ans. 4 a?. (170,1.) 2. (4 a 2 ) I 



3. (8m 8 )*. 4. (36m 4 )*. 5. (81 

6. (32m 10 )*. 7. (-125W 9 )*. 8. V4QcF*. 



9. VM^. 10. V27?/ 6 . 11. V6256W 

12. (4c 2 )l Ans. (2c)*. (170,1.) 13. (9 a 6 )*. 

14. (8 ft 3 )*. 15. (-27m 9 )*. 16. 



17. 9 4 . 18. IG 19. 

20. V28. Ans. 2V7. (170,2.) 21. VIS. 

22. (90)1 23. V45. 24. V72. 

25. V48. 26. (27)*. 27. V24. 

28 3 V8. 29. V56. 30. (96)*. 



EXERCISE IN SIMPLIFYING RADICAL QUANTITIES 261 



31. \/250. 
34. (320)1 
36. (27 



39. 

42 

44. 5(500)*. 



46. 



49. 2V256. 



52. 



55. VSOa'V. 



57. . 



60. 2(f)*. 

62. (f)i 

64. (})*. 

67. 9 -- 3 ' 



'V 



70 



73. 



76. 



"VgOa'v "'' 



\'a + 6 
//*.-3\i 



32. 2-V/81. 

35. (18 2 6) 

37. (75)* 
40. 



'162. 

i 



33. 

. 3 a (2 

38. V24T 2 . 
41. (98 c 2 )*. 
43. 2(84)*. 



45. 4 (24 a 4 



47. 



50. -V/1D2 a 2 . 



48. 
51. 

53. S/300 a 4 - 54. 

56. (1)1 Ans. 1(3)*. ( 170, 3.) 

58. (f)i 59. (1)1 

'. ( 170, 3.) 61. 

i / 5 

63. (To-) 3 - SUGGESTION. I 

\2 2 3 2x3 

65. VfV 66> "V^f- 

/T nr 

Vra' 



2 x 32 



68 



69. 



71. 



w . 
74. 




276 




75. 3(f)i 



/ . 9 

" ~ 



78. 



a 



262 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

172. Miscellaneous Exercise in Simplifying Radical Expressions. 
1. V80. 2. V288. 3. 2(175)1 ' 4. VlOO tfy 6 . 

5. v'aW. 6. (36)* 7. A/289. 8. 

9. ^64a. 10. -N/r 11. 4V27 a 3 6 5 . 12. ^v 



/3000 






13. \- - 14. -V^ - 15. V8--206 2 . 16. 3(18a?-9)* 
* a- a V 3 6 2 

Ans. to Ex. 15. 2 V2 5 6 2 . How is it obtained ? 

17. Describe the three kinds of simplifications we are using. 

18. Make rules for solving each kind. 



19. 3645. 20. Sm 21. 

22. V5 ar - 10 xy + 5 if. 23. V432 a 3 6 5 . 24. V50 ab' 2 c 2 



28. f--1 29. -[--] . 30. 

b Sa* cb 



31. . 32. . 33. 




34. V(a 2 -6 2 )(a + &). 35. V189. 36. vVfrV 4 ". 

"a 3 " . 38. 



(xyy b\d J 



40. a- ~ 41. 



.. , 97\i- 

43. ( - ] 44. 3a 2 A / 648a 71 . 45. V196 x 98. 

or 



ADDITION AND SUBTRACTION OF RADICALS 263 

46. V75 or 5 - 100 x s . 47. V4 a 3 - 24 a?x + 36 



I 2 a 
. \/- 
\2-- 



W625 z 
48. \/- - 49. 



a-f-36 



50 . . 51. 

\ 512 a 6 (a 

173. Putting Coefficients of Radical Quantities under the Root Sign. 

It is sometimes desirable to reverse the operation of the last 
articles, and, instead of removing factors from under the root sign, 
to put coefficients back into the radicand. 

1. 3(5)*. SOLUTION. 3(5)* = 9* x 5* = (45)*. ( 168, 1.) 



2. 4(3)1 3. 5VT4. 4. 6a(36)s 5. a 

6. 1(4)1 ^n*. (!)*=(*. 7. 4(3)* 8. T * r 



9. 3aVH. 10. 2SX5. 11. T 12. aV6. 



17. Which is greater, 3 V7 or 4 V4? 3V3or2V6? 3S/2or2\/7? 

174. Addition and Subtraction of Radical Quantities. When 
radical quantities are simplified, as explained in 170, they can 
be added and subtracted by adding and subtracting their coeffi- 
cients, provided the quantities are similar, that is, have the same 
radicand and same root index ( 23). 

1. Simplify 3 V45 + V20 7 V5 ; and check answer to two 
decimal places. 

SOLUTION CHECKING 

3(45)*= 9(5)* ( 170, 2) 3V45= 20.12+ 



+ (20) = + 2 (5) ( 170, 2) V20 = 4.47 

-7 (5)* = -7 (6)* _ -7V5 = --15.65 

Sum= 45)^ = 8.94+. 8 - 94+ - Check. 



264 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 



REMARK. The student may not have seen the object in simplifying 
radicals as explained in 170, while studying that article, but he ought to 
see it now. How much easier it is to calculate 4 V5 than to calculate 

3\/45+ V20-7 V5~! 

Simplify and check to two decimal places the eight following 
problems. The checking here serves two purposes : it checks the 
accuracy of the radical reduction, and also of the root extractions. 
If desired, check those problems which contain letters by assign- 
ing values to the letters. 



2. V27 + V12 + V75. 
4. 50* + S* - 98*. ' 
6. 600* -32-* + 108*. 



3. 8V125 + 2V80-3V45. 



5. 44 -2 



7. 250* - 16* - 54*. 



9- 



10. 2V3--1V12 + 4V27. 11. 40* + 25* + 2 (625)* 

12. 12 (32)* + 5 (162)* -512*. 13. ^192-7-^9 



14. 



15. 



17. 



3/T 



!b_ 
ac 




ab 



ax- 



by 2 




4 ax? 



by* 



20. 

T 

21. (4 a 

22. 3 



+ 6(75 a y + V3 a (a - 9 &) 2 . 
- ^/375^ + -v/54^. 



23. 



18 5 9 ' 

24. Make a rule for solving exercises in the addition and 
subtraction of radicals. How can the answers be checked? 



25. 



26. 



MULTIPLICATION AND DIVISION OF RADICALS 265 



tr. 

28. 



SUGGESTION. When simplified, are all the radicals similar ? See 29, 22. 

29. 5 V363~a 3 F- 3 V242 3 /. 

30. V2^-4o^-f-2a - V2 c^ + 4 00; + 2 a. 

SUGGESTION. Factor quantities under the radical signs. Ans. 2 V2 a. 






31. 

\ / 

*32. (5 a 5 + 30 a 4 + 45 a 3 ) * - V5 a 5 - 40 a 4 + 80 a 3 . 

/a -I- a? 



34. (a fljj-v^ 11 ^ a(a- 




35. ^p^ + -- Va 3 6 - 4 a 2 6 2 + 4 a& 3 . 

36. VHJ+3V8- V^O-iVMS-SV 11 ^"- V24f. 



37. 



38. 2 7 + 4 

SUGGESTION. Reduce decimal to common fraction. 

175. Multiplication and Division of Radicals having a Common 
Index. Before working, change problems given in the radical 
sign notation to the other, and then change the result back. 

1. Multiply 2 V 14 by 3 V21. 

SOLUTION. 2 (14)* x 3 (21)^ = 6 (14 x 21)^ ( 168, 1) = 6 (2 x 7 x 3 x 7)* 
CHECK. 7.48+ x 13.75 = 102.85. = 42 V6 = 102.88. 

Steps in the solution : (1. Multiplying coefficients for coefficient 
of product. 

(2. Multiplying radicands for radicand of product. Result 
merely indicated. 



266 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

(3. Separating each number in resulting radicand into its prime 
factors. 

(4. Simplifying the result as in 170. 

In division of radicals substitute the words " divide ' : for 
" multiply," and "quotient' 1 for "product" in the foregoing 
steps. 

2. 3(8)^x5(6)1 3. 2Vl5x5V27- 4. V20 x V30. 
5. 2(6)* x (18)1 6. V} X Vf 7. 4 Vf X V|. 

8. 8V70-iV63. Ans. 16 ()* = - 1 / VlO. 9. 3(2)^-2(3)*. 



10. (15)*-i-(f)* n. 2V6-SX|. 12. V21-S-VJ. 

13. 9 (7)* -i- 2 (21)* 14. 5Va^2V3o~. 15. (72)* -i- (81)* . 
16. (*)*x(|$)*. 17, aV^X&Va. 18. tycMftf + -\/cfl>y. 



19. A/n-iA/i. 20. V55 #?/ X V 66 



21. a6xcxca. 22. (x~yf X 

23. -\/168 x ^147. 24. (48)* -s- (56)* 

25. 7V75^5V28. 26. 6 (7)* -s- 126* 



27. ^a-j-Ca 28. 

29. Make a rule for the multiplication and division of mono- 
mials having a common root index. 

30. SXI6 x 5 v/4 x 3\/|. 31. ^343 x ^- 1568. 
32. VjxVJxV}. 33. 



34 (1 J y )^ -5- (5f)*. 35. J/3Tc X 



36. Va 6x Va 



37. (|m 8 )H-(|m) 38. A8a 2 - (2 a 






MULTIPLICATION AND DIVISION OF RADICALS 267 



39. Multiply 2V2 + V3by4V2--3 V3. 

SOLUTION CHECKING 

2(2)*+ 3* 4.56 

4 (2)* --3(3)* x .46 

8 x 2+4(6)* 2.0076 

_ 0(6)* - 3 x 3 

16 - 2(6)* - 9 = 7 -- 2 V6 = 2.1. 

40. (2V5- V7)(3V54V7). 41. [2(6)* -3(5)*] [3(3)* -f 2(5)*]. 
42. (L>V6-7V7)(4V34-2V5). 43. (4V6-3V745V8) x6V2. 
44. (7 + 3V7)(2V7-7). 45. (2a43V#)(3a-2 

46. (Vl5-V2043V21)x V5. 47. (VI5- V3) -5-V3. 
*48. (V6-2V?>4 V7)-f-V2. 49. (a-6Vc)(d-ev7). 

50. (or - ab V2 4 ^ 2 ) (o? 4 &6 V2 4 & 2 )' 

51. (2* 4- 3* -5*) (2* + 3* + 5*). 

52. [2(8)*4-3(5)*-7(2)*][72*-5(20)*-2(2)*]. 

53. riVi-Wi-f 



54. a?a?-- a?y 4- 



55. [a? 4 2/^][ Vx 4 




. Multiplication and Division of Radical Quantities which do 
not have a Common Index. Radical quantities that do not have 
a common index can be reduced so that they will have a common 
index, and can then be multiplied or divided. The operations 
depend on the principles of 168. 



268 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

i, 

SOLUTION 

(2 x 3)^ X ( 32 x 2)* = (2 x 3)6" x (3' 2 x 2)* ( 168, 2) 

= (2=2 x 3*)* x (3 6 x 23)^ ( 168, 3) 

= (2 2 x 3* x 36 x 2 3 ) ( 168, i) 

= (3 8 x 2 5 )* 

= 3(25 x 3*)* =3 #288. ( 170, 2) 

Steps in the solution : 

(1) Separate each of the radicands into its prime factors ( 49), and if 

not already so written, put in the fractional exponent notation. 

(2) Reduce the fractional exponents to a common denominator. 

(3) Raise each radicand to the power denoted by the numerator of its 

fractional exponent. 

(4) Change the product or quotient of the same roots to the like 

root of the product or quotient by converse o/ 168, 1. 

(5) Simplify the result by 170, and, if called for, change, back to 

the radical sign notation. 



2. V2 x \/4. 3. VI2 X ^14 4. ^I^F x V7 x. 

5. V6 x -v/9. 6. A 3 /4 X -\/6. 7. A/32 - 

8. 6-V2. 9. A/4 x-if -r- ^2^. 10. 

11. 10 -s- 6*. 12. 20-5-3(10)*. 13. 



14. +f 15. afn-ii. 18. 

17. -\/10x4-v / 2. 18. 2^x3^x5^ 19. 6(54) -3(2)i 

20. 2 S/:5 - 3 VL2. 21. V^-^-V^. 22. -x/iX^fWf. 

23. (5 + ^4 + 2 v / 5)( V 6+ V5). ( 35, 82.) 

24. 



RATIONALIZATION OF DENOMINATORS 269 

177. Rationalization of Denominators. (See 170, 4.) To ration- 
alize the denominator of a fraction is to cause the irrational 
quantity in its denominator to disappear. This is accomplished 
by multiplying both terms of the given fraction by some quantity. 

T 2 2 32 2(3)^ 2V3 

1. - SOLUTION. x - - = - 

V3 i i 3 3 



CHECK. A = _L_ = 1.155- ; = - = 1.155-. 

V3 1-732 3 3 

The advantage in this reduction consists in its replacing one root extrac- 
tion and one long division, by one root extraction and one short division. 



3 4 V2 _ V^ A Vaj- -1 

2. . 3. . 4. 5. 6. 

V7 V5 V7 V& 

7. - SUGGESTION. Multiply both terms by \/3. 



8 Q * in 10 11 12 

8. --9. - 10. 11. 



-tyx 2 3 A/25 5 V3 

13. Rationalize the denominator of = - = 

Vo+ V2 

SOLUTION. When the denominator is a binomial, the multiplier is found 
in a different way from that when the denominator is a monomial. 

In the case of a binomial denominator both terms of the given fraction 
are multiplied by the denominator with the sign between its terms changed. 
This multiplier is called the conjugate surd to the given denominator. 

32 5? _ 2^ 15^ - 6^ VT5 - Vd 
x = 



5-2 3 



n A ~. - ,. 

CHECK. -=.47+; = .47+ 

V5 + V2 

REMARK. Here by the rationalization we replace three root extractions 
and one long division by two root extractions and one short division. 



270 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

14 V3 + V2. 15 1 + 2V3. 3(5)1-2(2)* 

V3-V2 5--V3 



7*;-l 2-V3 5-2V5 

178. Powers and Roots of Radical Quantities. See 166. 

III. IMAGINARIES 

179. Imaginary Quantities. A pure imaginary is an even root of 
a negative number, as (- - 1) 2 . The sum of a real number and a 
pure imaginary, as 3 + 2 V- - 1, is called a complex number. 



180. Exercise. 1. (V- - 1) 2 = - 1, by the^definition ( 18). 
The letter i is often used to denote V--1. Then 2 = 1. 
Hence, i 3 = i 2 x i = * ; ^ 4 = r X *' 2 = 1 x 1 = -f- 1. 



2. Add 3 + 2 V : - 1 and 5 + 7 V : - 1. Ans. 8 + 9 V- 1. 
3. 



4. V : 25 + V36 = ? ^w*. lit. 170,2. 

5. V-4a 2 



6. 3V-20a V 80a=? ^4s. 2V- 



a. 



7. (6 + 2V--l)(3-4V- -1)=? 



9. (6 + 3) (7 + 4 S). 




^3 = V3 i 



10. (4-6V6) 2 . 11. 6V3^-2V5 = ? Ana. f V5. 



IV. EQUATIONS CONTAINING RADICALS 

181. Solution of Equations Involving Radicals. After putting 
one radical by itself on one side, apply power or root axioms. 

1. Solve x% = 27 for x. 
SOLUTION. (x 2 ) 3 = 27. 

x 1 ~ 3. (On extracting root of both sides denoted 

.. x = 9. by numerator of fractional exponent.) 



SOLUTION OF EQUATIONS CONTAINING RADICALS 271 



2. 0* = 8. 3. Vz = 4. 4. 4ajf = 

5. J 3^ = 25. 6 - \ / ? = 81. 7. oH==5. 

8 Vis - 4> 7 Q 

. ^/ I . i/. 

11. 3 Vie- 4 = 11. 

SOLUTION. 3 Vx = 15. (Transposing 4, Sub. Ax.) 

9 x = 225. (Squaring both sides, Power Ax.) 
x - 25. (Ax. ?) 

VERIFICATION. 3V25 4 = 11, or 15 - 4 = 11. 



12. l+Va = 5. 13. l+Va?--l=3. 



14. 8 V3a; 6 = 14. 15. 



16. a+3 = 2a;-10. 17. 

18. l + 2v / = 7--VaJ. 

SUGGESTION. Transpose Vx and combine it with 2 Vx. 

19. 2-\/x 3& = x--7^/x. 

SUGGESTION. After squaring, divide through by x. Then x = is a root. 



20. Vo + # -f- V# = 5. 

SOLUTION. V5~+le = 5-Vx. ( One radical quantity on one 

side. Axiom used ?) 

5 + x = 25 10 Vx + x. (On the right side we have the 

square of the difference of two 

quantities (5 and Vx) equals the 
square of the first, minus twice 
the product of the first by the 
second, plus the square of the 
second. What axiom is used ?) 
10 Vx = 20. ( Tne ra( ji ca i quantity put by 

itself on one side. Ax. ?) 

vx = 2- (Dividing equals by 10. Ax. ?) 

x - (Power Ax. Quote it and ex- 

plain how it is used.) 

VERIFICATION. V5 -f 4 + V4 = 5, or 3 + 2 = 5* 



272 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 



21. a? 4- 10 - - = 2. 22. 2 

23. o?* + 2 = (a; + 3l > )-. 24. (15 + x)? = 3(6)* - a?*. 

25. (a; + 25)* = 1 + 3*. 26 - ( + 13)* = 13 - oj*. 

SUGGESTION. Raise both members 
to the 3d power . 



3 _ 
27. v --1 = 2. 



28. -v/9 a; - 4 + 6 = 8. 29. 



30. V3~^ - A/3~+^~ 2 = 0. 31. 3+V-9a? = a?. 
32. -\/x + 6 = a. 33. Va# + 2 a6 a = 6. 



34. VaT+ar J --aj---J- = 0. 35. ?i = 9 - - Vn* + 9. 

36. a + 2-Vl(> + a; 2 =0. 37. V* - 4 + 3 = V* + 11. 



38. V4 a; 2 + 6 a; + 36=2 z+4. 39. V2(l - a;) (3 - 2 a,-) +1=2 
40. (4a? l -7aj+l)*=2-lf 41. (3 x- 5)* + (3 x + 7)* = 6. 
42. 



SOLUTION. 4 a + x = 4(6 + a) + 4 V(6 + *) + * (Ax.?) 



+ a; 2 = 4 & + 4 a - 4 a. (Ax.?) 



x 2 = 6 + x - a. (Ax.?) 

bx + x 2 = 6 2 + x 2 + a 2 + 2 6x - 2 a& - 2 az. (Ax. ?. 137, 14.) 
2 ax -- bx = a 2 -2ab + b' 2 . (Ax. ?) 






43. Make a rule for solving radical equations covering the 
cases where there are (1) One, (2) Two, and (3) Three radicals 
each containing the unknown. So long as the radical quantities 
containing x remain in the equation, what plan is followed ? 
(See suggestions above.) How is the equation solved after it is 

cleared of radicals ? 



44. V# a Vcu; + a 2 = Va, 45. aV# & = c V# d. 



HISTORICAL NOTES 273 

46. In the simple pendulum, (heavy weight hung by string, 
swung slightly back and forth) the time t in seconds equals very 
approximately TT times the square root of the length in feet, I, 

divided by the acceleration of gravity, g, or t = %/-. Solve this 
equation, first for I, and then for g. 

47. What is the length of a pendulum that vibrates in | sec- 
onds ? Use g = 32.2 and TT = 3.1410, substituting in the formula 
just found. 

48. Solve v = a \/-- for r. 49. Solve v = V- yh for h. 

\ r 



_ _ 
50. Solve = -y - for R. 51. Solve u = ~Vv- + 2/s for s. 

182. Historical Notes. In the very earliest stages of the de- 
velopment of algebra, as already stated in 43, all demonstrations 
and solutions were written out in full in words, so that an algebra 
looked like any other book, as a reader or history. Later, abbre- 
viations were introduced, but grammatical endings and agreements 
were still employed, just as students nowadays often write ='s. 
Later still, symbols were used freely. 

To show how the notation of algebra has changed we will give 
Diophantus's syncopated notation. Following the usual Greek 
notation for numbers, using a for 1, b for 2, c for 3, etc. (see 
203), he represented the unknown by a character somewhat like 
s (plural ss), its square by d y (for dynamis, power), its cube by c", 
known terms by m (for monad, unit), addition by juxtaposition, 
subtraction by /p, and equals by i. Coefficients were written after 
instead of before quantities.* 

Thus, c u d ss c v dy b m e i s a 

represents (4 x 3 + 3 x) (2 x- + 5) = x. 

* Instead of using the Greek letters, which are unfamiliar, the correspond- 



ing Roman letters are used. 



274 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 

As an example of a notation used by a writer of a time not long 
before our present symbols came into use we give that of Regio- 
montanus in a work prepared in 1464 and published in 1533. 

16 census et 2000 aequales 680 rebus, 
that is, 16 y? + 2000 = 680 things. 

Our present symbols, as already stated, began to be used in the 
sixteenth century. As showing in a convenient form how they 
gradually came into use, we give on the opposite page in tabular 
form a list of these symbols, with information about them. The 
references in the right column are to Chrystal's " Text-Book of 
Algebra," Ball's "A Short History of Mathematics," and Cajori's 
" A History of Elementary Mathematics." 

This table shows clearly how we came to duplicate symbols for 
the same operation. Thus, apposition, x , and all came to be 
used to denote multiplication. Similarly, the fraction sign, -;- , 
and / all denote division, and the radical sign and fractional ex- 
ponents both are used to indicate roots. Sir Isaac Newton, in a 
letter to Oldenburg of 13th June, 1676, writes : " Since algebraists 

-I o , 

write a 2 , a 3 , etc., for aa, aaa, etc., so I write a 2 , a 2 for Va, V 3 , 

etc. : and I write or 1 , or 2 , etc., for -, , etc." The radical sign, 

a act 

as the table shows, had been in use for over a hundred years before 
mathematicians thought of using fractional exponents. 

The symbol oo denotes an exceedingly large number, greater 
than any that can be named. It is often associated with 0, the 
being thought of as the limit of an exceedingly small number that 
differs from by a number less than any that can be assigned, 
however small. These two kinds of quantities are called infinites 
and infinitesimals, and are largely dealt with in a branch of math- 
ematics called the calculus, which was developed by Newton and 
Leibnitz about 1670. A quantity that is neither infinite nor in- 
finitesimal is called & finite quantity. The theorem on page 255, 
strictly speaking, should read, any finite quantity whose exponent 
is zero is equal to unity. 



HISTORICAL NOTES 



275 



TABLE 

SHOWING THE ORIGIN OF THE SYMBOLS IN COMMON USE 



SIGN, OR 
CONVENTION 


YEAR 
INTRO- 
DUCED 


GEN- 
ERAL 

USE 


BY WHOM 

INTRODUCED 


REMARKS 


Division. 
Dividend over 
Divisor, 


Very 
early, 




Hindus and Arabs, 


This is the oldest sign. 
Oblique line also used. 
See Ball, pp. 213, 148. 
Chrystal, p. 24. 


Apposition fur 
Multiplication, 


1200 






See Chrystal, p. 24. 


+ and - 


1525 


1630 


Rudolff and Stifel, 


Diff . of opin. as to orig. 
See Cajori, p. 141. 
Ball, p. 185. 


= 


1557 


1680 


Robert Recorde, 


Whetstone of "\Vitte. 
Ball, p. 191. 


v~ 


1551 
1633 


17th C. 


Rudolff, Stifel, 
Girard, 


V V= = 4thrt., VVV 
= cube rt., etc. 
Girard used indices as 
we do. 


Letters and 
Yinculum, 
Letters, 


1591 
1637 




Vieta, 
Descartes, 


Used capitals. Vowels 
for unknowns. 
Used small let. Last 
let. for unknowns. 


Parentheses, 


1629 


18th C. 


Girard, 


See Chrystal, p. 24. 


X, : : 


1631 


1686 


Oughtred, 


See Cajori, p. 234. 


x, >,< 


1631 




Harriott, 


Caj., p. 234, Ball, p. 213. 


-f- 


1630 




Pell, 


Ball, p. 214. 


Pos. Int. Exp's., 


1637 




Descartes. 


In analytic geometry. 


Dot for Mult., 


1(537 


18th C. 


Descartes, 


Ball, p. 213, C., p. 194. 


Frac. Exp's., oc , 


1658 




Wallis and Newton 


See Chrystal, p. 201. 


; 


1760 




Clairaut, 


Ball. p. 213. 


Asso. Com. and 
Distrib. Laws, 


1840 




Peacock, Gregory, 
and De Morgan, 


Cajori, p. 244. 


=,>, < 


Recent 






Ball, p. 214. 






CHAPTER XI 
QUADRATIC EQUATIONS 



183. A quadratic equation is one which, when reduced to its 
simplest form, contains the second, but no higher, power of the 
unknown quantity ; or, a quadratic is an equation of the second 
degree (see 57). 

Quadratic equations containing but one unknown quantity are 
of two kinds, complete and incomplete. A complete quadratic 
equation contains both the second and first powers of the un- 
known, as 3 x 2 -f- 5 x -f- 7 = 0. An incomplete or pure quadratic 
contains only the second power of the unknown, as 6 x 2 24 = 0. 

I. INCOMPLETE OR PURE QUADRATICS 

184. Solution of Incomplete Quadratics. The type form of such 
equations is mx 2 = n. To reduce to this type form it may be 
necessary to clear of fractions, transpose, and combine terms con- 
taining $. 

3 _4s* 
~ 



SOLUTION. 9 z 2 - 27 = 4 x" 2 + 18. (Ax. ?) 

9 y? - 4 x 2 = 27 + 18. (Unknowns to left member; 
,, 2 45 knowns to right. Ax. ?) 

z 2 = 9. (Ax. ?) 

x = 3. (Root Ax.) 

VERIFICATION. 9 3 = (36 + 18) *- 9, or 6 = 6. 

a. It is not necessary to write x 3, for x = 3 has the same roots 
as x = 3. Show that this is true by writing -f x = -f 3, -f- x = 3, 
- x 4- 3, - x = 3, and multiplying the third and fourth equations 
through by - 1. 

276 



APPLICATIONS OF INCOMPLETE QUADRATICS 277 



2. 7o 2 -2o=5a- 2 + 73. 3. 9a 2 - 53 = 6^ + 94. 

4. (o? + 4; 2 = 8a; + 25. 5. 3ar J + 12 = 7 2 -88. 

6. (3-7)(3a + 7)=32. 7. (. + 1) (a? - 1) = T V 

8. .c + 2# + 3=2a,- 2 + 5a. 9. a? + 3aj- 3= 66- 2x 2 . 



- t , or - 

a c a c 



10. aa; 2 b = C.JT + d. ^4/is. \- or - - V ; -> c^Z --be cd. 

* 



a # __ a& - 3 a a; -f 5 6 

11. -j 1/2. --- - u. 

x a x x 5 b 3 a + 10 b 



13 , _ ., 14 a; 3 __ 

' ' x-2 x 2 



15. Make a rule for solving incomplete quadratic equations, 
stating (1) What is done when fractions appear in the equation. 
(2) What the next step is. (3) How the terms are collected. 
(4) What the last step is. 



_ 



x a 2 ,x + a? ^ x + a x a 

=1. 19. 



0; 



2 . 

a 2 a? a- a; + 6 x b x 2 b 2 



20. V25 -- 6 a; + V25 + 6 x = 8. 21. V + a= 



22. 205^+3 + 2+2=1. 23. 

185. Applications of the Solution of Incomplete Quadratics. 

1. The area of a circle is 3.1416 r 2 . What is the radius of a 
circle (to three decimal places) whose area is 30.1 sq. ft. ? 

2. The surface of a sphere is 4 x 3.1416 r 2 . What is the radius 
of a sphere whose surface is 91 sq. meters ? 

3. Solve s = \gt 2 for t, reducing the answer to the form re- 
quired by 170, 3. 



278 QUADRATIC EQUATIONS 

4. In the formula s = gt 2 , s is the distance a body falls, t 
the number of seconds it takes, and g, the acceleration of gravity, 
is 32.2 ft. Find by substituting in the answer to the preced- 
ing exercise the number of seconds it takes a body to fall 64.4 ft. 

5. The volume of a cylinder is .7854 d 2 x I in which d is the 
diameter of one end and I is its length. Find to three decimal 
places the diameter of the end of a cylinder 14 inches long which 
contains 155 cu. inches. 

6. If d and D are the diameters of the pistons of a hydro- 
static press (see dictionary for description) and p and P are the 
pressures respectively on the small and large pistons, it is shown 
in physics that 

cl L_ 
P D 2 ' 

Solve this equation for D, having p, P, and d given. Solve also 
for d when p, P, and D are given. 

Find by substitution in the answer just obtained what D is 
when p = 2 lb., P= 150 lb., and d = 1 in. 



mv 2 



7. Solve /= for v. 



r 



8. Solve /= p for t. Simplify result. ( 170.) 

t 

V 2< 3 

9. Solve^=- for V and d respectively. Simplify. 

8 



10. How many rods of fence will inclose a square garden 
whose area is 2^ acres ? a acres ? 

11. The diagonal of a square is 16 ft. Find one side of it. 

12. Find the diameter of a down-spout for a roof 40 X 20 ft., 
using 1 sq. in. of pipe area for every 150 sq. ft. of roof area. 

13. Let p and p' be the periods of revolution of two planets 

r> 2 d 3 

about the sun, and d and d' their mean distances. Then E- = . 

p' 2 d' 3 

Solve this equation for p and also p', simplifying as in 170. 



SOLUTION OF QUADRATICS BY FACTORING 279 

14. What is the diameter of a pipe that will carry the same flow 
as four 1-inch pipes ? 

15. A side of an equilateral triangle is 8 ft. Find altitude x. 

Generalize by putting a for 8, and - for 4. 



16. The side of an equilateral triangle is 12 8 

ft.. Find its altitude by substituting in for- 
mula of Ex. 15. Find also its area. Find the 
side when the altitude is 20 inches. 



8 



4 

17. A regular hexagon is formed out of six 



> ^ 'I i til J_L VJ v Cl> ^ y 



equilateral triangles having a common vertex at its center. Find 
one side x of the hexagon if its area is 15 sq. ft. Draw diagram. 

18. One side of a square whose four corners fall on the circum- 
ference of a circle is 2 ft. Find its radius. Generalize. 

19. Radius of circle is 12 rods. Find side of inscribed (ver- 
tices on circumference) equilateral triangle. Generalize. 

SUG. Geometry shows distance from center to one side = radius -4- 2. 

II. COMPLETE QUADRATICS. SOLUTION BY FACTORING 

186. Solution of Quadratics by Factoring To Construct a Quad- 
ratic, its Roots being given. Properties of Quadratic Equations. 

Whenever the roots of a quadratic are either whole numbers or 
simple rational fractions, the factoring method ( 57) is usually the 
easiest and quickest. 

1. Solve and verify in x- -f 8 x = 33. 

SOLUTION, x 2 + 8 x 33 = 0, (Ax. ? Eight member made 0.) 
or, (x + 11 ) (x 3) = 0. (On factoring.) 

.-. x + 11 = 0, whence, x = 11, (On setting each factor equal to zero.) 
and x 3 = 0, whence, x = 3. 

VERIFICATION. (-- II) 2 + 8 x 11 = 33; 3 2 + 8 x 3 = 33. 

2. x 2 --6x = 16. 3. x 2 -f 12 x = 64. 4. x 2 + 1 7 x = 0. 
5. 2o?-5x = 3. 6. 6x 2 + l = 5a;. 7. 8a? 



280 QUADRATIC EQUATIONS 

8. Construct the equation whose roots are 3 and 4. 
SOLUTION. [* - 3] [x - ( - 4) ] = 0, 

or, y? + x 12 0. Ans. 

PROOF. (x 3) (x + 4) = 0. (On factoring.) 

.. x 3 = 0, whence, x 3, 
and x + 4 = 0, whence, x - 4. 

Evidently to form an equation ivhose roots are two given numbers, 
one subtracts each from x, multiplies the remainders together, and 
sets the product equal to zero. 

9. Form the equations whose roots are : 4 and 6 ; 2 and 5 ; 

11. 2 . 1 . _ 3. JL 
2", ^, ^, 4 , 5, 3. 



71. Q ft . 1 1 ^ 90 ft. 

'> J- > -*> - ) 1U, 10, ZU, - - D, 

10. Form the equation whose roots are a and b. 

11. In the equation or 2 -- 7 # + 12 = 0, what are the two roots ? 
What coefficient of the given equation is their product ? What 
coefficient with its sign changed is their sum ? 

12. Examining 11 and- 10 above we see that if the coefficient of 
3? in a quadratic equation is unity and the right member is zero, 
the coefficient of x with its sign changed equals the sum of the two 

j and the term which does not contain x equals their product. 



III. "COMPLETING THE SQUARE" METHOD 

187. Solution of Complete Quadratics by Completing the Square. 

Whenever it is not easy to solve an equation by factoring, it can 
be solved by a process called " completing the square." (See 117.) 

To complete the square, we think of the well-known expression 
a 2 + 2 ab + 6 2 , the square of a + b, and consider that we have the 
first two terms given to find the third. (See 116.) 

To find 6 2 from a 2 + 2 ab, divide the second term by twice the 
square root of the first term and square the resuU. 

Thus, 2 ab -r- 2 Vtr' = b ; and squaring 6 gives b 2 . 



COMPLETING THE SQUARE 281 

1. Solve and verify in the equation x 2 +- 14 x +- 45 = 0. 

SOLUTION. 

z 2 -f 14 x = 45. (Sub. Ax., known quan. to right side.) 

We have 14 x + 2 Vz 5 = 7 ; 7 2 = 49. Then, 

a;2 + 14 x _j_ 49 - 49 _ 45 - 4. (Add. Ax., 49 added to each side of 

the preceding equation.) 
x + 7 = 2. (Root Ax.) 

x = 7 + 2, or - 7 2, i.e. 5 or 9. ^4s. (Sub. Ax.) 
VERIFICATION 



(-5) 2 + 14 x --6 + 45 =0 

25 - 70 + 45 = 

0=0 



(-9) 2 + 14 x -9 + 45 = 

81 -- 126 + 45 = 

= 



2. z 2 -6z-91 = 0. 3. x 2 - 

4. aj"-4a;-77 = 0. 5. x- 2 +- 60 a? + 891 = 0. 

6. a* + 8 a; = 768. 7. 23 = 120-0* 

8. 24 = 11 a; -a 2 . 9. 225 = 30 x -x 2 . 

10. oj* _ 341 = 20 a. 11. x = -34. 

x 

SOLUTION. 3 x' 2 = 57 - - 10 . (On clearing of fractions, Ax. ?) 

3 x 2 + 10 x = 57. (Unknowns to left side, Ax. ?) 

The coefficient of x 2 must now be made a perfect square like a 2 in the for- 
mula p. 280. 

This is accomplished by multiplying the equation by 3. (It might be done 
by dividing through by 3, but this would introduce fractions into the solution, 
and it is usually better to avoid fractions.) 

9 a 2 + 30 ae= 171. (Ax. ?) 



We now complete the square. Thus, 30 x -f- 2 V9 x 2 = 5 ; 5- = 25. Then, 

9 x 2 + 30 x + 25 =: 196. (Ax. ?) 

3x + 5 =: 14. (Root Ax.) 

Bx= - 5 14- (Sub. Ax.) 

x = 3, or L 9 . (Div. Ax.) 

VERIFICATION. 3 = ^ - - -^ ; also - - ^ = - - 3 . 

3 



282 QUADRATIC EQUATIONS 

12. 2or J -7a + 5 = 0. 13. 9or 2 + 4z = 5. 14. 7 g 2 -f 26 = - 19. 

15. a 2 -fa; = 32. 16. 2 2 -z = i. 17. igx = $ x*. 

18. 8a 2 +-z = 30. 19. 3^ + 35 = 220;. 20. 18 a?- 27 a; == 26. 

21. 18 a,- 2 + 27 x = 26. 22. 3x* + 10 = 17x. 23. (2 a? - 3) 2 = 8 x. 

24. 2aj-3 = 3aj + 2aj-3. 25. a; + l(2a? + 3= 4 a; 8 -22. 



ofi 1 _ 27 = 

4z+-7 : "oj + 7* 7-5"2aj-13 

9. ^-4-.^ 

28. 



aj 



2 3 a; + l --! 



2 ,+l = 13 " 31 a; a; 

" 



6 aj + 1 ^' + 4 

32. Solve the preceding by the factoring method, and compare 
the length and difficulty of the two solutions. It will be found 
that some are easier by one method and some by the other. 

33. 5x 2 + 7x + l = 0. 

SOLUTION. 100 x' 2 -f 140 x = 20. (Mult, and Sub. Axs.) 

100 x 2 -f 140 x + 49 = 29. (Completing square, Ax. ?) 

lOz + 7 = V29. (Ax. ?) 



10 10 



= _. 1615 , or _ L2385+. 



VEBIPIOATIO*. 5 7 ' + 7 



29 - 14 V29 + 49 + 14 \/29 - 98 + 20 = 0. 

(On canceling and clearing.) 

a. In the verification only the positive root was used. That of the nega- 
tive root - x/21) 7 would differ from the other only in certain signs. 



COMPLETING THE SQUARE 283 

Solve and verify in the following as in the model just given : 
34. a^ + 3:c+l = 0. 35. 3xr-2x = 2. 36. 2^-7 






37. x*-9x+12=0. 38. 7tf + l = 7x. 39. 3ar'-ll 

40. 0^ 4^ + 7 = 0. See 179. 41. 2^5 #+4 = 0. 

42. Make a rule for the solution of quadratic equations. 

(1) What is the first step when fractions appear in the given 
equation ? 

(2) What terms are transposed and where ? 

(3) How are the terms collected ? 

(4) What is the equation multiplied or divided through by (when 
necessary) ? Ans. By some number which will make the first 
term a perfect square and positive, and at the same time make 
the subsequent part of the solution as short as possible. See 
b, p. 284. 

(5) How is the square completed ? 

(6) What is the next step ? 

(7) The next ? 

(8) The last ? (Roots containing irrationals should be calculated 
to three decimal places.) 

(9) How is the answer verified ? In verification use radical form 
of the answer, as in Ex. 33. 

Give the axiom used for each step except the third and (9). 

What it is best to multiply or divide through by is learned 
from practice. Solve and verify in the following, using the 
suggestions : 

43. In 7 tf + 21 x - 70= divide through by 7. 

44. In 3 x 2 + 2 x - 33 = multiply through by 3. 

45. In 2 x 2 17 x + 21 = multiply through by 8. 

46. In 15 x 2 + 5 x 70 = divide through by 15. 

47. In 18 x 2 - 12 x - 48 = divide through by 2. 



284 



QUADRATIC EQUATIONS 



48. In 18 x 2 - 17 x - 111 = multiply through by 72. 

b. What is called the Hindu rule for solving quadratics avoids fractions. 
It is, Multiply the equation through by 4 times the coefficient o/x' 2 . This was 
done in the third and last of the equations just given. But it is not always 
necessary to multiply by so large a multiplier to avoid fractions. Thus, frac- 
tions were avoided in Ex. 44. 



14 , 

+ - -, = x-l. 



9 2x-3 
#+22 4__9a? 6 



3 






Ix 



15 



-5#- V2-7?/ = 



49. 



51. 



53. 



55. 
56. 

57. ax 2 + bx + c = 0. 
SOLUTION. ax 2 + bx 

4 a' 2 2 + 4 abx 

4 a 2 x 2 + 4 aftx + & 2 

2 ax + b 



ert 
50. - 



3x 



52. 



4 o ^ 3 x 

-2 x + 1 



X ~r~ O X O O 



9 






^ 
" 



c, 

4 ac, 

b' 2 - 4 ac, 



2 4 ac, 



x 



6 V6- -- 4c 
- 



(Ax. ?) 

(Hindu rule, Ax. ?) 
(Comp. Sq., Ax. ?) 

(Ax.?) 

(Axioms ?) 



Tr 

VERIFICATION. 



a 



. , 
+ b 



A 
+ c = 0. 



- 4 ac - 2 b Vfr 2 - 4 ac + b 2 
4 a' 2 



V62._4 ac _fr2 4a C ^ 



c. Careless pupils sometimes want to put b V- 4 ac, for VJfl - 4 ac, or 
the like. Now the square of a monomial, as 3 a, is a monomial, and the 
square of a binomial, as a + 5, is a trinomial. Hence, a binomial, as 
b' 2 4 ac, cannot be the square of a rational quantity. The square root of 
such binomials can only be indicated. See 29, 22. 



COMPLETING THE SQUARE 



285 



58. mx 2 + nx + p = 0. 
60. tf-3bx = 



62. x 

64. 12x*-cx-2Q<? = 0. 

66. 9 a 4 b V - 6 a s b s x = b\ 

68. x 2 



70. 



a a; 



a 2 --x* 



59. 
61. 

63. 15^-2 

65. 356 2 = 9 

67. (m n) x 2 - - nx = m. 

69. 

71. T- 



4 4- 



72. V# 4- a V# a = V2 x. 

73. a^ 



74. ax 2 + bx + c = c!x 2 + 

SOLUTION, ax' 2 dx 2 + bx hx = -- c, (Transposing, Ax. ?) 

(a d)x 2 + (b h~) x = c, (Factoring.) 

4 (a d) 2 x 2 + 4 (a tZ) (& - h) x = - 4 ac + 4 cd (Hindu rale.) 



Omitting intermediate steps, x = 



- 



2 (a d) 

d. The unknowns are transposed to left member, all the x 2 terms coming 
first, followed by all the x terms. Then, x 2 is taken out of the x 2 terms and 
x out of the x terms. From this point the solution is as in other problems. 



7 ^ 1*~ I /"/ /y rt r* i //> 

I C/ */ Lt'tC' -- \Ai\j \jvU 

77. aj2--4aaj--10a; = --40a. 

70 a _ ^ 

I C7 . 






76. mx 2 -f- HO; = pic 2 - - g. 
78. cic 2 2 cdx = ax 2 cd 2 . 

80. 



a; a 



4 



lo 



81. 
83, 

85. 



82. 2s4- 



1 

t 



84. V = 



86. c Vl c 2 = n 



286 QUADRATIC EQUATIONS 

Solve and verify in the following : 
87. 3aj 2 + 8 + 3=0. 88. 5y a - 

89. 2 a; 2 - 11 a? + 16 = ( 179). 90. 4^-12^ 
91. 9aj 2 -6a?-2 = 0. 92. 



1 111 

93. - = ---- 1 (After simplifying, divide out by a 6.) 

a b -\-x a b x 

IV. THE FORMULA METHOD 

188. Solution of Quadratic Equations by Means of a Formula 

We solve the equation ax 1 + bx + c = 0, regarding it as a type of 
any or every quadratic, getting (see Ex. 57, 187) 



2a 

Now, having a given quadratic to solve, we compare its coeffi- 
cients with those of ax 2 -f bx + c = 0, thus getting values for a, b, 
and c. These values for a, b, and c are then substituted in the 
value of x just given. The result simplified is the answer sought. 

The student should practice solving the equation ax 2 -t-bx+c=Q 
until he is sure he can remember the process. One should know 
how to derive the formula before using it. 

1. Solve 2X 2 9 a + 10 = by the formula method. 

SOLUTION. Writing the given equation underneath the type form, we have 

ax 2 + bx + c = 0, 
2x*-Qx + 10 = 0. 

We see now that in this particular problem a = 2, b = 9, c = 10. 
Substituting these values for a, 6, and c in 

- 6 Vb' 2 4 ac 

jj 

2a 



_C_9) V(- 9Y- 4 x 2 x 10 
we get a = ^ '- ! 



or x- - -80_01_5 2 

4 4 2' 

VERIFICATION. 2 x 2 2 - 9 x 2 + 10 = ; 2(|) 2 - 9 X f + 10 = 0. 



SOLUTION BY MEANS OF A FORMULA 287 

Solve and verify in the following, using the model above : 

12. o2_7a;_|_6 = 0. 3. a? + 8x-2Q = 0. 

4. 6jf 2 -3oaj-6=0. 5. 2 a? + 13 a; + 15 =0. 

6. 6^ + 37 a? + 56 = 0. 7. 4ar>- 4a?- 3 = 0. 



8. + = <. 9. 



934 2* + l oj-3 

2x-\-l 5 #-8 

--- = - 

1-2* 7 2 

SUGGESTION. Equations 8, 9, 10 must be first cleared of fractions and 
then the terms transposed so that they have the form ax 2 + bx + c = 0. 

11. Make a rule for solving quadratic equations by the formula 
method. 

(1) Jn what form is the given equation put if it is not already in 

that form ? 

(2) How are the values of a, b, c in the formula obtained ? 

(3) What is done with these values ? 

(4) How are the two roots now found ? 

12. 5oj 2 -4i=12. 13. 2^ + 7a;-4 = 0. 



14. ar + 2_ 

SUGGESTION. Put in the formula 1 for a, 2p for 6, 3 q for c. 

15. 4ar-6f?.r = 4rf 2 . 16. G or + 8 m 2 = 49 ma. 



- rv 9 I -| -| *? -I O - y\ 99 7OO y!7 9 T 9 

17. or + 11 w = 12 ??i#. 1 8. rr.r crar 4 ctoaj = cr 0. 

SUGGESTION. Put in the formula d~ 6 2 for , --4 ab for 6, and -- a' 2 + 6 2 
for c. 

19. Solve problems 2-10 by factoring, and also by completing 
the square, and compare with the formula solutions. 

20. Solve maj 2 4-wa?+p = 0. Then, from the value of a? found 
make a rule for solving an equation of the form mx 2 -\- nx +p = 0. 



~ 

Hence, 



288 QUADRATIC EQUATIONS 

V. PROBLEMS 

189. Problems involving the Solution of Quadratic Equations. 

1. A man bought a quantity of cloth for $120. If he had 
bought 6 yd. more for the same sum, the price per yard would 
have been $ 1 less. What was the price ? 

SOLUTION. Let x = number of dollars in cost of one yard, 
Then, - = number of yards bought. 

Qu 

x -- 1 = number of dollars in supposed price. 

= number of yards from supposed price. 

x 1 

120 120 _ 6 

x -- 1 x 
120 x - 120 x + 120 = 6z 2 - Gx. (Mult. Ax.) 

6 x 2 -Gx- 120 = 0. (Sub. and Mult. Axs.) 

x i - x 20 = 0. (Div. Ax.) 

(x 5) (x + 4) = 0. (Factoring.) 

.-. x 5 = 0, or x = 5, Ans. 

x + 4 = 0, or x = 4. 

a. The second answer, 4, can have no meaning in this problem, since 
the nature of the problem is such that negatives are inadmissible. Let us 
substitute for the given problem another which has the same numbers in it 
and gives rise to the same equation, but which admits of the entrance of 
negatives. Thus : 

The exchange account of a banker amounted in a certain number of days 
to $120, during which time exchange remained the same for each day. Had 
the period differed from what it was by 6 days and in his favor (that is 6 more 
if premiums or 6 less if discounts), it would have made a difference of\per 
day. What was his daily premium or daily discount ? 

To this problem 4 is as satisfactory an answer as + 5. The 4 means 
a daily discount of 



General Note. The solution of quadratics by factoring is not feasible 
when the roots are irrational or imaginary. When the roots are rational, as 
they commonly are in problems and exercises, the factoring method (review 
for it 52, 57) generally gives the quickest and easiest solution, even 
when the coefficients are large. When it fails, one of the others must be 
chosen. The formula method is shorter, but harder to remember than that 
of 187. 






PROBLEMS INVOLVING QUADRATIC EQUATIONS 289 

2. Find a number such that if you subtract it from 10 and 
multiply the remainder by the number itself the product is 27. 

SOLUTION. Let x = number. Then, (10 x)x = 27, or, 2 10 x = - 27. 

a: 2 10 x + 25 = 2. (On completing the square.) 

x-5 = V~^2. (Ax, ?) 

x = 5 V^2. (Ax. ?) 
VERIFICATION. [10 (5 V^2) ] (5 V^) = 27. 



&. The imaginary answer 5 V 2 ( 179) shows that there is no real 
number which being multiplied by the remainder obtained from subtracting 
it from 10 gives 27. Thus, trying 8, we get 16 for product ; trying 7, we get 
21 ; trying 6, we get 24 ; trying 5, we get 25 ; trying 4, we get 24. Neither 
will any negative number satisfy the conditions. To make this problem 
"real" it must be stated as follows: What complex number on Argand's 
diagram is it which multiplied by the difference between 10 and itself gives as 
product 27 ? 

c. Algebra is a formal science made to cover all cases, and without any 
reference to particular problems. In problems which by their nature admit 
of negative numbers, a negative answer has meaning ; in others negative 
answers are inadmissible. Moreover imaginary values for the unknown in 
ordinary problems denote that the conditions are impossible of fulfillment. 

3. A railway train traveled 5 mi. an hour slower than usual, 
and was 1 hr. late in making a run of 280 mi. How many miles 
per hour did it travel? Solve first, and later generalize; solve 
and test generalized answer. Let a = number miles per hour 
slower, b = number hours late, c = number miles in whole dis- 
tance. See 123. 

4. A company of gentlemen engaged a supper and agreed to 
pay $ 80 for it. Four of the gentlemen failed to attend, and each 
of the rest paid $1 more than he expected to pay. How many 
were present at the supper ? Generalize. 

5. A man worked a certain number of days for $ 30. If he 
had received $1 a day less than he did, he would have been 
obliged to work 5 days longer to earn the same sum. How many 
days did he work ? Generalize. 



290 QUADRATIC EQUATIONS 

6. A picture that was 8 in. by 12 in. was placed in a frame of 
uniform width. If the area of the frame was e^ual to that of the 
picture, what was the width of the frame ? Draw diagram. 
Generalize. Check. 

7. Two men, A and B, can do a piece of work in 20 da. If B 
requires 9 da. more than A to do the whole work, how many days 
would it take each to do the whole work ? Generalize. 

8. A body of soldiers can form a hollow square 4 men deep. 
By re-forming the company with 36 men less on each outer side, 
the soldiers can form a solid square. Find the number of soldiers 
on one side of the solid square and the total number of soldiers. 

9. The perimeter of a rectangular field is 180 ft. and its area 
is 1800 sq. ft. Find its dimensions. Generalize. Check. 

10. The two digits of a number differ by 1 ; and if the square 
of the number be added to the square of the number with its 
digits reversed, the sum is 585. Find the number. 

11. Find the price of eggs when 2 less for 30 / raises the price 
a dozen. Generalize. Check. 



12. A crew can row downstream 18 mi. and back again in 7|- hr. 
Their rate upstream is 1|- mi. an hour less than the rate of the 
stream. Find the rate of the stream and of the crew in still water. 

13. A boatman rowed 8 mi. upstream and back in 3 hr. If 
the velocity of the current was 2 mi. an hour, what was the rate 
of rowing in still water ? Generalize. Check. 

14. Divide a straight line a inches in length into two parts so 
that the ratio of the whole line to the larger part may equal the 
ratio of the larger part to the smaller part. This was called by 
the Greeks dividing the line in "golden section." 

15. The circumference of the fore wheel of a coach is 5 ft. less 
than that of the hind wheel. If the fore wheel makes 150 more 
revolutions than the hind wheel in going a mile, what is the cir- 
cumference of each wheel ? Generalize. 



PROBLEMS INVOLVING QUADRATIC EQUATIONS 291 

16. Two points start out together from the vertex of a right angle 
along its respective sides, the one moving m ft. per second, the 
other n ft. per second. How long will they require to be c ft. apart ? 

17. The hypotenuse of a certain right-angled triangle is 8 ft. 
greater than the perpendicular, and 4 ft. greater than the base. 
What are the lengths of its sides ? Generalize. Check. 

18. Find two numbers whose difference is d and product is p. 
Verify answers. 

19. The product of two numbers is p and their quotient is q. 
What are the numbers ? Verify. 

20. It is required to find three numbers such that the product 
of the first and second equals a, the product of the first and third 
equals 6, and the sum of the squares of the second and third 
equals c. Verify answers. 

21. A farmer sowed one year a hectoliter of wheat; the next 
year he sowed what he harvested the first year less b hectoliters, 
and reaped c fold of what he sowed and d hectoliters besides. 
Assuming a like fruitfulness both years, what did he reap the 
first year ? 

SUGGESTION. Let x = number of hectoliters reaped first year. Then, he 
reaped second year x times the number of hectoliters he sowed ? Why ? 

22. A set out from C towards D and traveled 7 mi. per day. 
After he had gone 32 mi., B set out from D towards C and went 
every day -jL of the whole journey, and after he had traveled as 
many days as he went miles in a day, he met A. Required the 
distance from C to D. Verify. 

23. In a school field meet one runner won over another in the 
100-yd. dash by half a second, and his speed was .4 yd. greater 
each second than his competitor's. What was the time of each ? 

SOLUTION. Let x = number of seconds the winner took. 

x 4- .5 = number of seconds the other runner took. 



Then, -- - = .4, whence a? + .5 x = 125, giving x = 10.933+. 
x x + .5 

VERIFICATION. 100 -- 10.933 = 9.146; 100 -- 11.433 = 8.744; 9.146-8.744 
= .4, nearly. 



292 



QUADRATIC EQUATIONS 



24. From New York to Chicago over the New York Central 
and Lake Shore lines is 960 mi. The 20th Century Limited 
covers the distance in 5} hr. less time than the Fast Mail 
Limited going at a speed which is 12.9 mi. an hour faster. What 
is the rate and time of each train ? 

SUGGESTION. After clearing and transposing make the coefficient of a; 2 
unity by dividing through by its coefficient. Keep three or four decimal 
places. 

25. In the Vanderbilt cup race for automobiles on Long 
Island, in 1905, Hemery on a French Darracq beat Tracy on a 
Locomobile by 22 min., the total distance covered being 283 mi. 
Hemery's rate was 41 mi. an hour faster than Tracy's. What 

was Hemery's rate ? 

26. A man proposes to cut 3-inch squares 
out of the four corners of a piece of tin 
2 inches longer than wide, turning up the 
sides so as to make a box containing 72 
cu. in. How long shall he cut the piece of 
tin? 

27. An ancient problem from India reads as follows : One 
fourth of a herd of camels was seen in the forest; twice the 
square root of the number had gone on mountain slopes ; and 15 
remained on bank of the river. How many camels were there ? 

28. The dimensions of a rectangular 
garden are 3 a and 4 a. Running along 
three sides of it, as in the marginal figure, 
is a walk of the same width throughout, 
whose area is b. What is the width of 
the walk ? 





x 



30 



4 a 



29. Suppose a 12 ft. and b = 448 sq. ft. Find by substituting 
in the formula just derived the width of the walk. Again, sup- 
pose that x= .5 rod and b = 2 sq. rods, and find the corresponding 
value of a. 



SIMULTANEOUS QUADRATICS 293 

30. The length AB of a rectangle, ABCD, exceeds its width 
AD by 119 ft., and the diagonal, BD joining two opposite vertices, 
equals 221 ft. Find AD and AB. Draw diagram. 

31. A circular basin is surrounded by a path 6 ft. wide, and the 
area of the path is -J of the area of the basin. Find the radius of 
the basin. 

32. E. A. Waltbour rode a bicycle in a mile race against time 
in 1904, making a record. Had he ridden the mile in 2.2 seconds 
less time, he would have covered 2.74 ft. more each second. 
What was his time to the nearest tenth of a second ? 

33. The distance from Chicago to San Francisco over the 
Northwestern, Union Pacific, and Southern Pacific railroads 
is 2280 mi. Discarding small fractions, the Overland Limited 
covers the distance in 15^ hr. less time than the China and 
Japan Fast Mail traveling 5 miles an hour faster. What is the 
time of each train ? 

34. In the Olympic games at Athens, April, 1906, the Mara- 
thon race of 42 kilometers (nearly 26 mi.) was won by the 
Canadian Sherring. Had he covered the distance in 6 min. less 
time, he would have had to increase his rate .5334 kilometer 
per hour. What was his time ? 

VI. SIMULTANEOUS QUADRATICS 

190. Simultaneous Quadratics. (See 98.) Degree of an Equa- 
tion. ( 57.) Simultaneous quadratic equations may be sepa- 
rated into two classes : those in which there is only one quadratic 
equation ; and those in which there are two or more quadratics. 

Tlie degree of an equation is the greatest exponent or sum of 
exponents of the unknown quantities in any one term. Notice 
that by this definition xy = is a quadratic equation as well as 



We may think of the x 2 , xy, and y 2 terms as denoting areas, 
while x and y terms denote merely lengths. 



294 



QUADRATIC EQUATIONS 



191. Simultaneous Equations one of which is Quadratic can 

always be solved by the Substitution Method. (See 99.) 



. (1) 



(2) 



9Q 4. ii 

SOLUTION. (2i) x = - ^. (Sub. and Div. Axs.) 

o 



^54, (99.) 



or. 



841 - 232 y + 16 ya 87y-12yg ,. 
9 3 



then, 20 f- - 29 y = 355, (Mult, and Sub. Axs.) 

whence, solving, y = 5, or |. 

On substituting these values of y in (2), x = 3, or - 7 3 2 -. 



VERIFICATION. (1) 9 + 3x3x5=54. 
(2) 3x3 + 4 x5 = 29. 



5184 
~~~ 



3x 



20 



2. 



4. 



6. 



8. 



10. 



f (1) xy x = 0, 



1( 



1 a? y = 5. 



12. 



xy = 45, 
a; y = 4. 

x y 



3. 



5. 



7. 



9. 



11. 



f (1) 9 
(2) 3x 



1 



-2y = 



9 t 9 -i orv 

ry- + a- = loO, 
s - q = 8. 



13. 



?/ o; 



QUADRATIC EQUATIONS SOLVED BY SQUARING 295 



14. 



16. 



a+^= 



~ y - iL. 

x 2 

r , 3s 



o 
3_ 
12 r ' 3s 



-j 

+ = 2. 

' o 



15. 



17. 



7 



236 



a; 



x 



_ _ 40 

a? y 21 

2 = 1. 



18. 



20. 



t 29 



-t = -2. 



[_ 4 xii = a 2 b 2 . 



19. 



32 15 

a?-3 = 1. 



r 



22. 



23. 



o? 2 -f- dy 2 = q. 



24 



26. 



1 

; +?/ = 5 



25. 



27. 



= . 



192. Quadratic Equations Solved by Squaring, Adding or Subtract- 
ing, and Extracting the Square Root. Problems in which one equa- 
tion is quadratic and the other simple (as in the preceding article), 
or in which both are quadratic, can be solved more elegantly by 
the following method, provided their coefficients are so related 
that the sum and difference of two quantities can be found. 



296 QUADRATIC EQUATIONS 

i I (1) a 2 + f = 250, 

1(2) x-y=4. 

SOLUTION 

(2i) x' 2 2ry-\- y' 2 = 16 (Squaring Ax.) 

(3) 2xy =234 (Sub. Ax.) 
(1) a; 2 + ?/ = 250 

(4) x 2 + 2 xy + y 2 = 484 (Add. Ax.) 
(4 X ) + ?/ = 22 (Ax.?) 

02) x - y = 4 

2x =26 or -- 18, (Ax. ?) 

a; = 13 or - 9. (Ax. ?) 

2y = 18 or -- 26, (Ax. ?) 

y = 9 or -- 13. (Ax. ?) 

VERIFICATION. 13 2 + 9 2 = 250, 
13 -9 = 4. 



(_. 13)2 = 250, 
( _ 9 ) -(-13) = 4. 



2. 3. 



4. i 5. - 

= 36. I a? + y = 8. 



r ^ 2 _i_ ^ ? / _i_ ? /2 _. g^ SUGGESTION. From (1) subtract (2) 

8 squared. Then get (3) xy = 18. Add (3) 

~y~ to (1), and extract square root. 



* = 21 
9. 10. 



QUADRATIC EQUATIONS SOLVED BY SQUARING 297 






(2) 4 a? + ^ = 25. 



SUGGESTION. Square (1) and subtract 
(2). Now to get 2 x y find the middle 
term of the trinomial square whose end 
terms are 4x 2 +# 2 (116, 20). Subtract 
4 xy = 24 from (2). Knowing the values of 
2 x + y and 2 x y it is easy to find x and ?/. 






15. 



9 2? 



= 241. 



16. 



18. 



(1) a 2 + a# + y 2 = 13, 



(2) x-y = 2. 



20 



. { 



17 



\i x 



19 



21. 



. I 



4 ? 2 = 21. 



22. 



24. 



26. 



(9x>- -13ajy + 9/ = 101, 

{ 3 xy = 12. 

r y? + xy 4. f = a , 



= 6. 



23. 



25. 



27. 



or xy -\-y 2 = 19. 



[ icy = 



a 



a; 4- y = a, 
4 a? = a 2 6 2 . 



28. 



1-1 = 12, 

a? y 



SUGGESTION. Solve as in Ex. 1 
without clearing either equation of 
fractions. (See 100, 25.) 



29. 



36 ' 



30. 



- + - = 8, 

a y 

i + 1 + 1 , = 49. 

a?" 0$ ?/" 



298 APPLICATIONS OF ALGEBRA IN SCIENCE 

31. The course of a yacht 30 mi. in length is in the shape of a 
right triangle one leg of which is 2 mi. longer than the other. 
What is distance along each side ? 

32. A can fold 3600 bills in 54 min. less time than B, and both 
can fold 7200 bills in 4 hours. How many can each fold in one 
hour ? 

33. H and K run a race of 1 mi., K winning by 1 min. If H 
increases his speed 2 mi. an hour, and K lessens his by the same 
amount, H will win by 1 min. Find former rates of each. 

NOTE. Certain equations of higher degrees, as x 4 13 x 2 + 36 = 0, or, 
2/ 6 + 7 y 3 8 = 0, can be solved as quadratics. 

ALGEBRA USED IN TRIGONOMETRY 

193. Exercise in solving Formulas in Trigonometry. In the fol- 
lowing s stands for a number called the sine of an angle, c for the 
cosine, t for the tangent, k for the cotangent, and z for the secant 
of the angle. 

o 

1. Solve t = - f or s and c. 

c 

2. Solve t = - for k. 

k 

3. Solve s 2 + c 2 = 1 for s and c. 
4 Solve 2 + 1 = z 2 for t and z. 

o 

5. Given t - and s 2 -f- c 2 = 1, to find t in terms of s. 

c 

6. Given c = -, and t 2 + 1 = z 2 , to find c in terms of t. 

z 

7. Given s = , to find first a and then h. 

h 

8. Given t = -, to find first a and then b. 

b 

9. Given a 2 = 6 2 + c 2 2 6m, to solve for m. 



INDEX* 



(The numbers refer to pages.) 



ABSCISSA, 174. 

Addition, of algebraic numbers, 12-19. 

of equations, 157. 

of fractional exponent quantities, 251. 

of fractions, 101. 

of imaginaries, 270. 

of literal quantities, 30. 
Aggregation, symbols of, 28, 40, 44. 
Algebra, 9, 12, 122. 
Antecedent, 117. 
Approximation, 228. 
Arithmetic, 9, 12, 138, 220. 
Arithmetical numbers, 12. 
Arrangement of letters, 51. 
Associative law, 24, 25. 
Average, 228. 
Axes, 173. 
Axioms, 123, 124. 

BINOMIAL, any power of, 204. 

cube of, 70. 

factoring, 76, 194. 

square of, 66, 68. 
Binomial coefficients, 204. 
Binomial quantity, 27. 
Biographical notes, 200. 
Brace, bracket, 28. 

CHECK, 13, 33. 
Circle, 234, 235. 
Clearing of fractions, 125. 
Coefficient, 27. 

of radical, 257. 
Common factor, 98. 
Common multiple, 92. 
Commutative law, 24, 25. 
Comparison, elimination, 160. 
Complete quadratic, 276, 279, 280, 286. 
Completing square, 280. 



Complex fraction, 114. 

Complex number, 270. 

Conditional equations, 122. 

Conjugate surd, 269. 

Consequent in proportion, 117. 

Consistent equations, 183. 

Continuation symbols, 12. 

Coordinates, 174. 

Cost problems generalized, 228. 

Cube, 26. 

Cube root, 213-218. 

DECOMPOSITION of a fraction, 107. 
Degree, of a term, 27. 

of an equation, 90, 293. 
Develop, 205. 
Difference, 15. 
Distributive law, ?5-, 
Divisibility, 72, 'li. 
Division, 23, 52. 

axiom of, 123. 

See Multiplication. 
Domestic science problems, 246-249. 

ELIMINATION, 154. 

by addition and subtraction, 157. 

by comparison, 160. 

by substitution, 154. 

in quadratics, 294-297. 

special methods, 162, 169, 

three unknowns, 168. 
Equal, 61, 141. 
Equation, 60, 61, 90, 122. 

balance, 127. 

compatible, 183. 

conditional, 122. 

construction of, 280. 

graphical solutions, 182. 

identical, 122. 

incompatible, 184. 



* This can be used to advantage in reviews by having students ignore all topics the page 
numbers of which are greater than the last page studied by the class. 

299 



300 



INDEX 



Equation, in physics, 241. 

linear, 18]. 

literal, 188. 

quadratic, 90, 186, 276-298. 

radical, 270. 

simple, 90, 123. 

simultaneous, 154, 293. 

systems of, 102, 184. 
Equivalent equations, 162. 

fractions, 102. 
Errors in algebraic operations, 115. 

of calculation, 228. 
Evolution, 207. 
Exercise, 131. 
Expand, expansion, 204. 
Exponent, 26. 

in division, 52, 251. 

in multiplication, 45, 251. 

zero and negative, 254. 
Exponents, 219. 

Expression, algebraic = quantity. 
Extremes, 117. 

FACTORS, 74, 194. 

common, 98. 

in multiplication, 21, 45. 

order of, 51. 

prime, 74, 89. 

zero, 46. 

Figures, 9, 12, 166. 
Formulas, 66, 72, 204, 242, 246. 
Fraction, 96, 251. 

proper, 107. 
Function, 132. 
Fundamental Laws, 24. 

GENERALIZATION, 193. 
Geometry in algebra, 230. 
Graphs, 119, 172-184, 236. 

HIGHEST common factor, 98. 
Hindu rule, 284. 
Historical notes, 64, 199, 273. 
Hypothesis = what is given. 

IDENTICAL equation, 122. 
Imaginaries, 270, 289. 
Improper fraction, 107. 
Incomplete quadratic, 276. 
Indeterminate equations, 193. 



Index, 26, 257. 
Infinity, infinitesimal, 274. 
Interest, 222. 
Involution, 202. 
Irrationals, 257. 

LAWS, 24, 28, 250-256. 

Leading letter, 51. 

Letters, 51, 125. 

Like signs, 22. 

Literal equations, 188-192. 

Literal quantity, 27, 219. 

Locus, 180. 

Lowest common multiple, 92. 

MANUAL training problems, 240. 
Mean, Arithmetical, 228. 
Means, in proportion, 117. 
Mechanics, formulas of, 118, 273, 278 
Member of an equation, 61. 
Minus, 12-19. 
Mixed quantity, 101. 
Monomial, 27. 
Multiple, 92. 
Multiplication, axiom, 124. 

fractions, 109. 

imaginaries, 270. 

literal quantities, 45, 250, 251. 

radicals, 265-268. 

simple algebra numbers, 21, 25. 

NAUGHT. See Zero. 
Negative exponent, 254. 
Negative quantities, 12, 19. 
Negative series, 12, 16. 
Negative solutions, 193, 
Newton's binomial theorem, 204 
Notation, 9-65. 

radicals, 257. 
Numbers, algebraic, 12. 

complex, 270. 

irrational, 257. 

natural or arithmetical, 12. 

rational, 257. 

real, 289. 
Numerical value, 28, 245. 

OMISSION symbols, 12. 
Ordinate, 174. 
Origin, 173. 



INDEX 



301 



PARENTHESIS, 28, 40, 44. 
Percentage, 220. 
Physics, 240, 244, 273, 278. 
Plotting points, 175. 
Plus, 12, 19. 
Polynomial, 27. 

factoring, 197. 

square of, 207. 
Positive, 12, 19. 
Power, 26, 50, 202. 

of quantity, 50, 111. 

of radicals, 270. 
Precedence of operations, 28. 
Prime factor, 89. 
Prime marks, 190. 
Prime quantities, 74, 89. 
Principles, 24. 
Problem, 131. 

Proofs, 55, 77, 117, 125, 254. 
Proper fraction, 107. 
Property, of identical equations, 122. 

of quadratic equations, 280. 
Proportion, 117, 141. 
Pure quadratic, 276. 

QUADRANT, 173. 

Quadratic equations, 90, 186, 276. 

complete, 279. 

incomplete or pure, 276, 277. 

simultaneous, 293. 
Quadrinomial, 83, 19(>. 
Quantity, 26. 

known and unknown, 123. 

RADICALS, 257. 
Radicand, 257. 
Ratio, 9, 117. 
Rational numbers, 257. 
Rationalization of denominators, 269. 
Real quantity, 289. 
Reciprocal, 159. 
Reduction, of fractions, 96. 

of radicals, 259, 269. 
Root, of equation, 123. 

of quantity, 26, 50, 207. 

SATISFY an equation, 123. 
Scale, algebraic, 16. 
Science, 240, 244, 246, 273, 278. 
Series, 12. 



Signs, 13, 19, 21, 23. 
Similar radicals, 26;!. 
Similar terms, L'7. 
Simple equation, 90, 123. 
Simple term, 27. 
Simultaneous equations, 154. 

quadratic, 293. 

three unknowns, 168. 
Solution, 123. 

negative, 193. 
Square, 26. 
Square root, 209-213. 
Squared paper, 119, 172, 236. 
Substitution elimination, 154, 294. 
Subtraction, 15, 34. 
Surd, 257. 
Symbols, of aggregation, 28. 

of omission, 12. 

of operation, 13, 15, 25, 26. 

of quantity, 9. 
Symmetry, 218. 
System of equations, 162, 184. 

TERMS, 27. 

of a fraction, 96. 

of a proportion, 117. 

similar, 27. 

Theorems, 66-74, 194, 204, 207. 
Transposition, 61, 125. 
Trigonometrical expressions, 298. 
Trinomial, 27. 

factoring, 79, 194. 

square of, 207. 

UNITY, as coefficient, 27. 

as denominator of integral quantity, 
105. 

as exponent, 26. 
Unlike signs, 22, 23. 

terms, 27. 

VALUE, numerical, 12, 28. 
Verification, 123. 
Vinculum, 28. 



ZERO, as exponent, 254. 
as factor, 46. 



PLANE GEOMETRY DEVELOPED 
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Bv EUGENE RANDOLPH SMITH, A. M. : Head of 



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ADAMS'S PHYSICS 

By CHARLES F. ADAMS, A. M., Head of the Depart- 
ment of Physics, Central High School, Detroit. 



Physics for Secondary Schools (Text-book) . . . $1.20 
New Physical Laboratory Manual 60 



THE text-book meets the demands of all college entrance 
boards, and of the New York State Syllabus. It is 
strong in the theory of physics, and is very thorough, 
treating of a great number of different phenomena, and 
explaining the reasons for their existence. 
^| The book is particularly teachable. The language is so 
simple and clear, and the illustrations are so numerous and 
illuminating, that the average pupil will encounter no difficulty. 
Throughout the controlling thought has been to make the 
work interesting and practical, as well as to afford the best 
mental discipline. 

^j The problems are unusually numerous, thus giving the 
teacher a wide choice. They emphasize and illustrate the princi- 
ples involved, the numbers being chosen so that the actual 
arithmetical work shall be easy. Wherever possible the funda- 
mental principles are enforced and brought home to the pupil 
by illustrations touching his daily life. The lecture table 
experiments have been selected only after a great deal of thought 
and care. Several are wholly new and particularly interesting. 
^[ The Laboratory Manual embodies the results of twelve 
years' experience in conducting laboratory work in physics. 
The 78 exercises are all simple, and the directions for manip- 
ulation clear. The College Entrance Requirements and the 
New York State Syllabus are fully covered, and there is 
enough additional matter to enable any teacher to make out a 
course of work adapted to his particular needs. The Appendix- 
contains general directions for the use of apparatus, and 
twenty tables of formulas and physical constants. 



AMERICAN BOOK COMPANY 



PSYCHOLOGY AND PSYCHIC 

CULTURE 

By REUBEN POST HALLECK, M.A., Principal, 
Louisville Male High School 

1.25 



IN this text-book the study of psychology is made distinctly 
useful. The volume is brief in compass, clear in state- 
ment, and interesting in treatment, and elevating and 
inspiring in its influence on the mind, the life, and the charac- 
ter. Instead of being a mere bundle of abstractions, dry 
facts have been enlivened and elucidated by illustrations and 
anecdotes, making it more than commonly attractive. 
^| The coordinate branches, physiological and introspective 
psychology, receive their proper share of attention. The 
physical basis of psychology is fully recognized. The sub- 
ject is introduced by a chapter on the nervous mechanism at 
the disposal of the mind, and is effectively illustrated by dia- 
grams and figures of the brain, nerve-fibers, etc. 
^ The treatment is singularly clear and simple. The psychic 
truths are expressed in such a manner that students will have 
no difficulty in forming definite ideas regarding them. The 
work abounds in concrete illustrations and incidents so aptly 
applied that they deeply impress the mind, 
^j At the same time the order of treatment is natural and 
logical and in harmony with the practical aims and purpose 
of the work. Special attention is devoted to the cultivation of 
the mental faculties. The abstract laws of the mind are first 
explained and developed, and afterwards the established truths 
of the science are applied. The fact that a knowledge of the 
laws and processes of the mind should be applied to self-im- 
provement, and to the best methods of gaining that mental and 
moral equipment required for a successful struggle in the 
battle of life, is constantly impressed on the mind of the 
student. 



AMERICAN BOOK COMPANY 



CLARK'S 
THE GOVERNMENT 

WHAT IT IS, WHAT IT DOES 

#0.75 



THIS text-book furnishes a unique presentation of the 
subject, treating of the principles of general govern- 
ment before considering those of local government. 
Its method of treatment encourages independent thought 
and personal research. This appears, not only in the 
supplementary work at the end of each chapter, but also in 
the problems of government given at intervals. Maps and 
diagrams are used, besides suggestive illustrations to reinforce 
the text. 

^[ Following an account of the government in general the 
book treats of the chief functions of government in a definite 
and logical manner. It then explains the American system 
of central government and the local government by the 
people ; it discusses voting, State and national constitutions, 
and the relation of nation, State, county, township, and city 
to each other. The principal officials of the various govern- 
mental systems are taken up, and their duties, qualifications, 
etc., properly considered. 

^| The latter part of the book presents important chapters on 
certain practical operations of government, such as : Trials, 
law-making, party nominations, political issues, and political 
corruption. A short summary of the principles of inter- 
national law and of the commoner laws pertaining to business 
and property is also included. 

^[ The author lays emphasis on the importance of State 
governments, and enriches the text with illuminating com- 
parisons with foreign governments. He presents the subject 
as a science, a complete and sensible system of interdependent 
parts and defined limits, with a single object the good of 
the people. 



AMERICAN BOOK COMPANY 



HUNTER'S ELEMENTS OF 

BIOLOGY 

By GEORGE WILLIAM HUNTER, A. M., Instructor 
in Biology, De Witt Clinton High School, New York City. 

1.25 



THIS work presents such a correlation of botany, zool- 
ogy, and human physiology as constitutes an adequate 
first-year course in biology. The foundation principles, 
upon which the present correlation of subjects is made, are 
that the life processes of plants and of animals are similar, and 
in many respects identical ; that the properties and activities 
of protoplasm are the same whether in the cell of a plant or 
of an animal ; and that the human body is a delicate machine, 
built out of that same mysterious living matter, protoplasm. 
With such a foundation this correlation is simple and natural. 
^[ The course is designed to give to students a general con- 
ception of the wide range of forms in plant and animal life ; 
to lead them to observe the various processes carried on by 
plants and animals, and to study only so much of structure 
as is necessary for a clear comprehension of these processes ; 
and to help them to understand the general structure of the hu- 
man body, and the way to care for it. 

^[ The treatment follows the order in which the topics are 
likely to be taken up when the work is begun in the fall. 
The laboratory and field work is interesting and readily 
comprehended. The questions are few and simple; they 
apply to structures easily found, and deal with externals only. 
The experiments outlined in the book do not require an ex- 
tensive laboratory equipment. Excellent results may be ob- 
tained with little or no apparatus, except that made by the 
pupils and teacher working together. 

^[ The course combines in excellent proportion text-book 
study, laboratory experiments, field work, and work for 
oral recitation, and is attractive, accurate, and informative. 



AMERICAN BOOK COMPANY 

(168) 



MYTHOLOGIES 

By H. A. GUERBER 

Myths of Greece and Rome. Illustrated . . . . $1.50 
Myths of Northern Lands. Illustrated . . . . 1.50 
Legends of the Middle Ages. Illustrated .... i 50 



MYTHS of Greece and Rome is well adapted for gen- 
eral reading, but it is of particular value in connection 
with the study of the classics. So intimately are these 
myths connected with our civilization, and so great is their 
influence upon our literature and art, that they should be 
familiar to every person. As told here, the repulsive features 
of heathen mythology are omitted. Excellent reproductions 
of ancient and modern masterpieces of painting and sculpture 
are plentifully used to illustrate the volume. The closing 
chapter includes an analysis of the myths in the light of philo- 
logy and comparative mythology. 

^j The grim sense of humor and the dark thread of tragedy 
running through Northern mythology, and found in the re- 
ligion of no other race, have left their indelible imprint upon 
our literature. In Myths of Northern Lands, these myths are 
told without unnecessary comment, and in a charming man- 
ner which can not fail to hold the attention. As in the 
other volumes, one of the most interesting and valuable features 
is the large number of reproductions of works of art. 
^| The object of Legends of the Middle Ages is to familiarize 
students with the legends which form the principal subjects of 
mediaeval literature, and whose influence is everywhere ap- 
parent in the subsequent history of literature and art. In 
connection with the various legends, appropriate questions are 
given from mediaeval and modern writings, illustrating the 
style of the poem in which they are embodied, or lending 
additional force to some point in the story. 



AMERICAN BOOK COMPANY 



DESCRIPTIVE 

CATALOGUE OF HIGH 

SCHOOL AND COLLEGE 

TEXT-BOOKS 

Published Complete and in Sections 



WE issue a Catalogue of High School and College Text- 
Books, which we have tried to make as valuable and 
as useful to teachers as possible. In this catalogue 
are set forth briefly and clearly the scope and leading charac- 
teristics of each of our best text-books. In most cases there 
are also given testimonials from well-known teachers, which 
have been selected quite as much for their descriptive qualities 
as for their value as commendations. 

^j For the convenience of teachers this Catalogue is also 
published in separate sections treating of the various branches of 
study. These pamphlets are entitled : English, Mathematics, 
History and Political Science, Science, Modern Languages, 
Ancient Languages, and Philosophy and Education. 
^| In addition we have a single pamphlet devoted to Newest 
Books in every subject. 

^ Teachers seeking the newest and best books for their 
classes are invited to send for our Complete High School and 
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interest. 

^| Copies of our price lists, or of special circulars, in which 
these books are described at greater length than the space 
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^| All correspondence should be addressed to the nearest 
of the following offices of the company: New York, Cincin- 
nati, Chicago, Boston, Atlanta, San Francisco. 



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